Electric Potential — Competition Zone

If potential (in volts) in a region is expressed as V(x, y, z) = 6xy − y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :

1. $-\left(6\hat{\text i} + 5\hat{\text j} + 2\hat{\text k}\right)$
2. $-\left(2\hat{\text i} + 3\hat{\text j} + \hat{\text k}\right)$
3. $-\left(6\hat{\text i} + 9\hat{\text j} + \hat{\text k}\right)$
4. $-\left(3\hat{\text i} + 5\hat{\text j} + 3\hat{\text k}\right)$

Answer

$-\left(6\hat{\text i} + 5\hat{\text j} + 2\hat{\text k}\right)$

Reason —

Given, V(x, y, z) = 6xy − y + 2yz.

The electric field is the negative gradient of the potential,

$\vec{\text E} = -\left(\dfrac{\partial \text V}{\partial \text x}\hat{\text i} + \dfrac{\partial \text V}{\partial \text y}\hat{\text j} + \dfrac{\partial \text V}{\partial \text z}\hat{\text k}\right)$

Taking the partial derivatives,

$\dfrac{\partial \text V}{\partial \text x} = 6\text y, \qquad \dfrac{\partial \text V}{\partial \text y} = 6\text x - 1 + 2\text z, \qquad \dfrac{\partial \text V}{\partial \text z} = 2\text y$

Therefore

$\vec{\text E} = -\left[6\text y\ \hat{\text i} + (6\text x - 1 + 2\text z)\hat{\text j} + 2\text y\ \hat{\text k}\right]$

At the point (1, 1, 0),

$\vec{\text E} = -\left[6(1)\hat{\text i} + (6 - 1 + 0)\hat{\text j} + 2(1)\hat{\text k}\right] \\[1em] = -\left(6\hat{\text i} + 5\hat{\text j} + 2\hat{\text k}\right)$

The diagrams below show regions of equipotentials.

A positive charge q is moved from A to B in each diagram.

1. Maximum work is required to move the charge q in figure (iii).
2. In all the four cases the work done is the same.
3. Minimum work is required to move the charge q in figure (i).
4. Maximum work is required to move the charge q in figure (ii).

Answer

In all the four cases the work done is the same.

Reason — The work done in moving a charge between two points in an electrostatic field is

$\text W = \text q\ \Delta \text V$

and depends only on the potential difference between the two points, not on the shape or spacing of the equipotential surfaces or on the path followed. In each of the four diagrams the point A lies on the 10 V equipotential and the point B on the 40 V equipotential, so ΔV = 30 V in every case. Hence the work done is the same in all four cases.

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, −σ and +σ respectively. The potential of shell B is :

1. $\dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text b^2 - \text c^2}{\text b} + \text a\right]$
2. $\dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text b^2 - \text c^2}{\text c} + \text a\right]$
3. $\dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text a^2 - \text b^2}{\text a} + \text c\right]$
4. $\dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text a^2 - \text b^2}{\text b} + \text c\right]$

Answer

$\dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text a^2 - \text b^2}{\text b} + \text c\right]$

Reason —

Given, three concentric shells of radii a, b and c with surface charge densities + σ, − σ and + σ respectively.

The charges on the shells are

$\text Q_\text A = \sigma(4\pi\text a^2), \qquad \text Q_\text B = -\sigma(4\pi\text b^2), \qquad \text Q_\text C = \sigma(4\pi\text c^2)$

The potential at the shell B is the sum of the potential due to the inner shell A (which behaves as a point charge at the centre), the potential due to its own charge, and the potential due to the outer shell C (which is the same at all interior points),

$\text V_\text B = \dfrac{1}{4\pi\epsilon_0}\left[\dfrac{\text Q_\text A}{\text b} + \dfrac{\text Q_\text B}{\text b} + \dfrac{\text Q_\text C}{\text c}\right]$

Substituting the charges,

$= \dfrac{1}{4\pi\epsilon_0}\left[\dfrac{\sigma 4\pi\text a^2}{\text b} - \dfrac{\sigma 4\pi\text b^2}{\text b} + \dfrac{\sigma 4\pi\text c^2}{\text c}\right] \\[1em] = \dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text a^2}{\text b} - \text b + \text c\right] \\[1em] = \dfrac{\sigma}{\epsilon_0}\left[\dfrac{\text a^2 - \text b^2}{\text b} + \text c\right]$

The given graph shows variation (with distance r from centre) of :

1. electric field of a uniformly charged spherical shell
2. potential of a uniformly charged spherical shell
3. electric field of a uniformly charged sphere
4. potential of a uniformly charged sphere

Answer

potential of a uniformly charged spherical shell

Reason — The graph is flat from r = 0 up to r = r₀ and then falls off smoothly with increasing r.

Due to a charged spherical shell, the potential at any interior point is the same as that at its surface,

$\text V_{\text{inside}} = \text V_{\text{surface}} = \dfrac{\text{KQ}}{\text r_0}$

which explains the constant portion of the graph. Outside the shell,

$\text V_{\text{outside}} = \dfrac{\text{KQ}}{\text r} \quad \Rightarrow \quad \text V \propto \dfrac{1}{\text r}$

which explains the falling portion. The electric field of a shell would be zero inside and then drop as $\dfrac{1}{\text r^2}$, while for a uniformly charged solid sphere the potential rises to $\dfrac{3}{2}\dfrac{\text{KQ}}{\text R}$ at the centre and is not constant inside. Hence the graph represents the potential of a uniformly charged spherical shell.

A uniformly charged ring of radius 3a and total charge q is placed in XY-plane centred at origin. A point charge q is moving towards the ring along the Z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is :

1. $\sqrt{\dfrac{2}{\text m}}\left(\dfrac{1}{5}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$
2. $\sqrt{\dfrac{2}{\text m}}\left(\dfrac{2}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$
3. $\sqrt{\dfrac{2}{\text m}}\left(\dfrac{1}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$
4. $\sqrt{\dfrac{2}{\text m}}\left(\dfrac{4}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$

Answer

$\sqrt{\dfrac{2}{\text m}}\left(\dfrac{2}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$

Reason —

Given, radius of the ring = 3a, charge on the ring = q, moving charge = q of mass m, initial position z = 4a.

The potential on the axis of a charged ring at an axial distance z is

$\text V = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\sqrt{\text R^2 + \text z^2}}$

At z = 4a :

$\text V_1 = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\sqrt{(3\text a)^2 + (4\text a)^2}} \\[1em] = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{5\text a}$

At the centre (z = 0) :

$\text V_2 = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{3\text a}$

To just cross the origin, the whole of the kinetic energy must be used up in increasing the potential energy,

$\dfrac{1}{2}\text m\text v^2 = \text q(\text V_2 - \text V_1) \\[1em] = \dfrac{\text q^2}{4\pi\epsilon_0}\left(\dfrac{1}{3\text a} - \dfrac{1}{5\text a}\right) \\[1em] = \dfrac{\text q^2}{4\pi\epsilon_0}\cdot\dfrac{2}{15\text a}$

Therefore

$\text v^2 = \dfrac{2}{\text m}\left(\dfrac{2}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right) \\[1em] \Rightarrow \quad \text v = \sqrt{\dfrac{2}{\text m}}\left(\dfrac{2}{15}\dfrac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$

Four equal point charges Q each are placed in the XY-plane at (0, 2), (4, 2), (4, − 2) and (0, − 2). The work required to put a fifth charge Q at the origin of the coordinate system will be :

1. $\dfrac{\text Q^2}{4\pi\epsilon_0}$
2. $\dfrac{\text Q^2}{4\pi\epsilon_0}\left(1 + \dfrac{1}{\sqrt3}\right)$
3. $\dfrac{\text Q^2}{2\sqrt2 \pi\epsilon_0}$
4. $\dfrac{\text Q^2}{4\pi\epsilon_0}\left(1 + \dfrac{1}{\sqrt5}\right)$

Answer

$\dfrac{\text Q^2}{4\pi\epsilon_0}\left(1 + \dfrac{1}{\sqrt5}\right)$

Reason —

Given, four charges Q at (0, 2), (4, 2), (4, − 2) and (0, − 2).

The distances of these points from the origin are

$\text r_1 = 2, \qquad \text r_2 = 2, \qquad \text r_3 = \sqrt{4^2 + 2^2} = \sqrt{20}, \qquad \text r_4 = \sqrt{20}$

Since potential is a scalar, the potential at the origin is

$\text V = \text{KQ}\left[\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{\sqrt{20}} + \dfrac{1}{\sqrt{20}}\right] \\[1em] = \text{KQ}\left[1 + \dfrac{2}{2\sqrt5}\right] \\[1em] = \text{KQ}\left[1 + \dfrac{1}{\sqrt5}\right]$

The work required to bring the fifth charge Q from infinity to the origin is

$\text W = \text Q\text V \\[1em] = \dfrac{\text Q^2}{4\pi\epsilon_0}\left(1 + \dfrac{1}{\sqrt5}\right)$

A positive point charge is released from rest at a distance r₀ from a positive line charge with uniform density as shown in figure. The speed (v) of the point-charge, as a function of instantaneous distance r from line charge, is proportional to :

1. $\text v \propto \text e^{+\text r/\text r_0}$
2. $\text v \propto \log\left(\dfrac{\text r}{\text r_0}\right)$
3. $\text v \propto \left(\dfrac{\text r}{\text r_0}\right)$
4. $\text v \propto \sqrt{\log\left(\dfrac{\text r}{\text r_0}\right)}$

Answer

$\text v \propto \sqrt{\log\left(\dfrac{\text r}{\text r_0}\right)}$

Reason —

The electric field due to an infinitely long uniformly charged wire of linear charge density λ at a distance r is

$\text E = \dfrac{\lambda}{2\pi\epsilon_0\text r} \\[1em] = -\dfrac{\text{dV}}{\text{dr}}$

Integrating between r₀ and r,

$\Delta \text V = -\displaystyle\int_{\text r_0}^{\text r}\dfrac{\lambda}{2\pi\epsilon_0\text r}\ \text{dr} \\[1em] = -\dfrac{\lambda}{2\pi\epsilon_0}\log\left(\dfrac{\text r}{\text r_0}\right)$

The charge is released from rest, so the loss in potential energy appears as kinetic energy,

$\dfrac{1}{2}\text m\text v^2 = \text q(\text V_{\text r_0} - \text V_\text r) \\[1em] = \dfrac{\text q\lambda}{2\pi\epsilon_0}\log\left(\dfrac{\text r}{\text r_0}\right) \\[1em] \Rightarrow \quad \text v^2 \propto \log\left(\dfrac{\text r}{\text r_0}\right) \\[1em] \Rightarrow \quad \text v \propto \sqrt{\log\left(\dfrac{\text r}{\text r_0}\right)}$

An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10^-29 C-m. What is the potential energy of the electric dipole?

1. −9 × 10^-20 J
2. −10 × 10^-29 J
3. −20 × 10^-18 J
4. −7 × 10^-27 J

Answer

−7 × 10^-27 J

Reason —

Given,

• Electric field, E = 1000 V m^-1
• Angle, θ = 45°
• Dipole moment, p = 10^-29 C-m

The potential energy of a dipole in a uniform electric field is

$\text U = -\vec{\text p}\cdot\vec{\text E} \\[1em] = -\text p\text E\cos\theta$

Substituting the values,

$\text U = -(10^{-29})(1000)\cos 45° \\[1em] = -(10^{-26})(0.707) \\[1em] \text U \approx -7\times10^{-27}\ \text J$

There is uniform spherically symmetric surface charge density at a distance R₀ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed v[R(t)] of the distribution as a function of its instantaneous radius R(t) is :

Answer

The graph in which the speed increases from zero at R₀ and then flattens out, approaching a constant value as R(t) becomes large.

Reason —

Let the radius of the distribution at any instant t be R, the initial radius being R₀. By the conservation of energy,

$0 + \dfrac{\text{KQ}^2}{2\text R_0} = \dfrac{1}{2}\text m\text v^2 + \dfrac{\text{KQ}^2}{2\text R}$

Differentiating this equation with respect to R,

$\dfrac{1}{2}\text m(2\text v)\dfrac{\text{dv}}{\text{dR}} - \dfrac{\text{KQ}^2}{2\text R^2} = 0 \\[1em] \Rightarrow \quad \dfrac{\text{dv}}{\text{dR}} = \dfrac{\text{KQ}^2}{2\text{mv}\text R^2}$

Here $\dfrac{\text{dv}}{\text{dR}}$ is the slope of the v-R graph. It decreases as both v and R increase, and the slope tends to zero as R → ∞. Hence the curve starts at v = 0 when R = R₀, rises steeply at first and then becomes almost horizontal, the speed approaching a limiting value.

A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a would be :

1. $\dfrac{\text Q(\text a^2 + \text b^2 + \text c^2)}{4\pi\epsilon_0(\text a^3 + \text b^3 + \text c^3)}$
2. $\dfrac{\text Q(\text a + \text b + \text c)}{4\pi\epsilon_0(\text a^2 + \text b^2 + \text c^2)}$
3. $\dfrac{\text Q}{4\pi\epsilon_0(\text a + \text b + \text c)}$
4. $\dfrac{\text Q}{12\pi\epsilon_0}\cdot\dfrac{(\text{ab} + \text{bc} + \text{ca})}{\text{abc}}$

Answer

$\dfrac{\text Q(\text a + \text b + \text c)}{4\pi\epsilon_0(\text a^2 + \text b^2 + \text c^2)}$

Reason —

Given, three concentric shells of radii a, b and c with equal surface charge densities, carrying a total charge Q.

Since $\sigma_\text a = \sigma_\text b = \sigma_\text c$,

$\dfrac{\text Q_\text a}{4\pi\text a^2} = \dfrac{\text Q_\text b}{4\pi\text b^2} = \dfrac{\text Q_\text c}{4\pi\text c^2} \\[1em] \Rightarrow \quad \text Q_\text a : \text Q_\text b : \text Q_\text c = \text a^2 : \text b^2 : \text c^2$

Also Q_a + Q_b + Q_c = Q, so

$\text Q_\text a = \dfrac{\text a^2}{\text a^2 + \text b^2 + \text c^2}\text Q, \quad \text Q_\text b = \dfrac{\text b^2}{\text a^2 + \text b^2 + \text c^2}\text Q, \quad \text Q_\text c = \dfrac{\text c^2}{\text a^2 + \text b^2 + \text c^2}\text Q$

For a point with r < a, the potential due to each shell equals the potential at its own surface. Hence

$\text V = \dfrac{1}{4\pi\epsilon_0}\left[\dfrac{\text Q_\text a}{\text a} + \dfrac{\text Q_\text b}{\text b} + \dfrac{\text Q_\text c}{\text c}\right] \\[1em] = \dfrac{1}{4\pi\epsilon_0}\cdot\dfrac{\text Q(\text a + \text b + \text c)}{\text a^2 + \text b^2 + \text c^2}$

The electric field in a region is given by $\vec{\text E} = (\text A\text x + \text B)\hat{\text i}$ where E is in NC^-1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V₁ and that at x = − 5 is V₂, then V₁ − V₂ is :

1. −48 V
2. −520 V
3. 180 V
4. 320 V

Answer

180 V

Reason —

Given, $\vec{\text E} = (20\text x + 10)\hat{\text i}$ N C^-1.

The potential difference is obtained from

$\Delta \text V = -\displaystyle\int_{\text r_1}^{\text r_2}\vec{\text E}\cdot\text d\vec{\text r} \\[1em] \text V_1 - \text V_2 = -\displaystyle\int_{-5}^{1}(20\text x + 10)\ \text{dx}$

Integrating term by term,

$= -\left[\dfrac{20\text x^2}{2} + 10\text x\right]_{-5}^{1} \\[1em] = -\left[10\text x^2 + 10\text x\right]_{-5}^{1}$

Evaluating the limits,

$= -\left[\left(10(1)^2 + 10(1)\right) - \left(10(-5)^2 + 10(-5)\right)\right] \\[1em] = -\left[(10 + 10) - (250 - 50)\right] \\[1em] = -\left[20 - 200\right] \\[1em] = 180\ \text V$

A point dipole $\vec{\text p} = -\text p_0\hat{\text x}$ is kept at the origin. The potential and electric field due to this dipole on the Y-axis at a distance d are, respectively : (Take V = 0 at infinity)

1. $\dfrac{\left|\vec{\text p}\right|}{4\pi\epsilon_0\text d^2},\ \dfrac{-\vec{\text p}}{4\pi\epsilon_0\text d^3}$
2. $0,\ \dfrac{\vec{\text p}}{4\pi\epsilon_0\text d^3}$
3. $\dfrac{\left|\vec{\text p}\right|}{4\pi\epsilon_0\text d^2},\ \dfrac{\vec{\text p}}{4\pi\epsilon_0\text d^3}$
4. $0,\ \dfrac{-\vec{\text p}}{4\pi\epsilon_0\text d^3}$

Answer

$0,\ \dfrac{-\vec{\text p}}{4\pi\epsilon_0\text d^3}$

Reason —

Given, $\vec{\text p} = -\text p_0\hat{\text x}$ placed at the origin, and the field point A lies on the Y-axis at a distance d.

Since the dipole axis is along the X-axis, the point A on the Y-axis lies on the equatorial line of the dipole. Hence

$\text V_\text A = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text p\cos 90°}{\text d^2} \\[1em] = 0$

The electric field at a point on the equatorial line is directed antiparallel to the dipole moment,

$\vec{\text E}_\text A = -\dfrac{1}{4\pi\epsilon_0}\dfrac{\vec{\text p}}{\text d^3} \\[1em] = -\dfrac{1}{4\pi\epsilon_0}\dfrac{(-\text p_0\hat{\text x})}{\text d^3} \\[1em] = \dfrac{\text p_0}{4\pi\epsilon_0\text d^3}\hat{\text x}$

Note: The textbook answer key gives option 2 for this question, but the worked hint given in the textbook itself obtains $\vec{\text E}_\text A = \dfrac{\text p_0}{4\pi\epsilon_0\text d^3}\hat{\text x}$, which is option 4. The correct option is therefore 4.

A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of −4 Q, the new potential difference between the same two surfaces is :

1. V
2. 2V
3. −2V
4. 4V

Answer

V

Reason —

Let r₁ be the radius of the inner solid sphere and r₂ that of the outer surface of the shell, and let $\text k = \dfrac{1}{4\pi\epsilon_0}$.

In the first condition, the outer shell is uncharged, so both potentials are due to the charge Q alone,

$\Delta \text V = \text V_{\text{in}} - \text V_{\text{out}} \\[1em] = \dfrac{\text{kQ}}{\text r_1} - \dfrac{\text{kQ}}{\text r_2} \\[1em] = \text{kQ}\left(\dfrac{1}{\text r_1} - \dfrac{1}{\text r_2}\right) \\[1em] = \text V$

In the second condition, the shell carries a charge − 4Q as well, and this charge contributes the same amount $-\dfrac{4\text{kQ}}{\text r_2}$ to both the potentials,

$\Delta \text V' = \left(\dfrac{\text{kQ}}{\text r_1} - \dfrac{4\text{kQ}}{\text r_2}\right) - \left(\dfrac{\text{kQ}}{\text r_2} - \dfrac{4\text{kQ}}{\text r_2}\right) \\[1em] = \dfrac{\text{kQ}}{\text r_1} - \dfrac{\text{kQ}}{\text r_2} \\[1em] = \text V$

Hence the potential difference does not depend on the charge given to the outer shell and remains the same.

Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to :

1. 2 : 1
2. 1 : 2
3. 10 : 7
4. 5 : 7

Answer

2 : 1

Reason —

For a charge accelerated from rest through a potential difference V,

$\dfrac{1}{2}\text m\text v^2 = \text q\text V \\[1em] \Rightarrow \quad \text v = \sqrt{\dfrac{2\text q\text V}{\text m}}$

The hydrogen ion and the singly ionized helium atom carry the same charge, but the mass of the helium ion is four times that of the hydrogen ion. Hence

$\dfrac{\text v_\text H}{\text v_{\text{He}}} = \sqrt{\dfrac{\text q_\text H \text m_{\text{He}}}{\text m_\text H \text q_{\text{He}}}} \\[1em] = \sqrt{\dfrac{4\text m_\text H}{\text m_\text H}} \\[1em] = 2$

that is, the ratio of the final speeds is 2 : 1.

Concentric metallic hollow spheres of radii R and 4R hold charges Q₁ and Q₂ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) − V(4R) is :

1. $\dfrac{3\text Q_2}{4\pi\epsilon_0\text R}$
2. $\dfrac{3\text Q_1}{16\pi\epsilon_0\text R}$
3. $\dfrac{\text Q_2}{4\pi\epsilon_0\text R}$
4. $\dfrac{3\text Q_1}{4\pi\epsilon_0\text R}$

Answer

$\dfrac{3\text Q_1}{16\pi\epsilon_0\text R}$

Reason —

Given, spheres of radii R and 4R with charges Q₁ and Q₂, having equal surface charge densities.

$\dfrac{\text Q_1}{4\pi\text R^2} = \dfrac{\text Q_2}{4\pi(4\text R)^2} \\[1em] \Rightarrow \quad \text Q_2 = 16\text Q_1$

Writing $\text k = \dfrac{1}{4\pi\epsilon_0}$, the potentials at the two surfaces are

$\text V(\text R) = \dfrac{\text{kQ}_1}{\text R} + \dfrac{\text{kQ}_2}{4\text R}, \qquad \text V(4\text R) = \dfrac{\text{kQ}_1}{4\text R} + \dfrac{\text{kQ}_2}{4\text R}$

Therefore

$\text V(\text R) - \text V(4\text R) = \dfrac{\text{kQ}_1}{\text R} - \dfrac{\text{kQ}_1}{4\text R} \\[1em] = \dfrac{3}{4}\cdot\dfrac{\text{kQ}_1}{\text R} \\[1em] = \dfrac{3\text Q_1}{16\pi\epsilon_0\text R}$

Two point charges 4q and −q are fixed on the x-axis at $\text x = -\dfrac{\text d}{2}$ and $\text x = \dfrac{\text d}{2}$, respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :

1. decrease by $\dfrac{\text q^2}{4\pi\epsilon_0\text d}$
2. decrease by $\dfrac{4\text q^2}{3\pi\epsilon_0\text d}$
3. increase by $\dfrac{2\text q^2}{3\pi\epsilon_0\text d}$
4. increase by $\dfrac{3\text q^2}{4\pi\epsilon_0\text d}$

Answer

decrease by $\dfrac{4\text q^2}{3\pi\epsilon_0\text d}$

Reason —

Given, 4q at $\text x = -\dfrac{\text d}{2}$ and − q at $\text x = \dfrac{\text d}{2}$, and the charge q is taken from the origin to x = d.

Writing $\text k = \dfrac{1}{4\pi\epsilon_0}$ :

At the origin, the distance from each fixed charge is $\dfrac{\text d}{2}$,

$\text U_1 = \text{kq}\left[\dfrac{4\text q}{\text d/2} - \dfrac{\text q}{\text d/2}\right] \\[1em] = \text{kq}^2\left[\dfrac{8}{\text d} - \dfrac{2}{\text d}\right] \\[1em] = \dfrac{6\text{kq}^2}{\text d}$

At x = d, the distance from 4q is $\dfrac{3\text d}{2}$ and that from − q is $\dfrac{\text d}{2}$,

$\text U_2 = \text{kq}\left[\dfrac{4\text q}{3\text d/2} - \dfrac{\text q}{\text d/2}\right] \\[1em] = \text{kq}^2\left[\dfrac{8}{3\text d} - \dfrac{2}{\text d}\right] \\[1em] = \dfrac{2\text{kq}^2}{3\text d}$

The change in the energy of the charge is

$\Delta \text U = \text U_2 - \text U_1 \\[1em] = \dfrac{2\text{kq}^2}{3\text d} - \dfrac{6\text{kq}^2}{\text d} \\[1em] = -\dfrac{16\text{kq}^2}{3\text d}$

Since U₁ > U₂, there is a decrease, of magnitude

$\dfrac{16}{3\text d}\cdot\dfrac{\text q^2}{4\pi\epsilon_0} = \dfrac{4\text q^2}{3\pi\epsilon_0\text d}$

The path being a semicircle does not matter, since the electrostatic field is conservative.

Two identical electric point dipoles have dipole moments $\vec{\text p}_1 = \text p\hat{\text i}$ and $\vec{\text p}_2 = -\text p\hat{\text i}$ and are held on the x-axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is :

1. $\dfrac{\text p}{\text a}\sqrt{\dfrac{1}{2\pi\epsilon_0\text{ma}}}$
2. $\dfrac{\text p}{\text a}\sqrt{\dfrac{2}{\pi\epsilon_0\text{ma}}}$
3. $\dfrac{\text p}{\text a}\sqrt{\dfrac{1}{\pi\epsilon_0\text{ma}}}$
4. $\dfrac{\text p}{\text a}\sqrt{\dfrac{3}{2\pi\epsilon_0\text{ma}}}$

Answer

$\dfrac{\text p}{\text a}\sqrt{\dfrac{1}{2\pi\epsilon_0\text{ma}}}$

Reason —

The electric field due to the first dipole at an axial distance a is

$\vec{\text E} = \dfrac{2\vec{\text p}}{4\pi\epsilon_0\text a^3}$

The second dipole is located in this field, so its potential energy is

$\text U = -\vec{\text p}_2\cdot\vec{\text E} \\[1em] = -(-\text p)\left(\dfrac{2\text p}{4\pi\epsilon_0\text a^3}\right) \\[1em] = \dfrac{2\text p^2}{4\pi\epsilon_0\text a^3}$

When the dipoles are infinitely far apart, U' = 0. Hence the loss in potential energy appears as the kinetic energy of the two dipoles,

$\Delta \text U = \text U - \text U' \\[1em] = \dfrac{2\text p^2}{4\pi\epsilon_0\text a^3} \\[1em] = 2 \times \dfrac{1}{2}\text m\text v^2 \\[1em] \text m\text v^2 = \dfrac{2\text p^2}{4\pi\epsilon_0\text a^3} \\[1em] = \dfrac{\text p^2}{2\pi\epsilon_0\text a^3} \\[1em] \text v^2 = \dfrac{\text p^2}{2\pi\epsilon_0\text{ma}^3} \\[1em] \Rightarrow \quad \text v = \dfrac{\text p}{\text a}\sqrt{\dfrac{1}{2\pi\epsilon_0\text{ma}}}$

A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r) as shown in figure. If the surface charge densities on the two shells are equal, the electric potential at the common centre is:

1. $\dfrac{1}{4\pi\epsilon_0}\dfrac{(\text R + \text r)}{(\text R^2 + \text r^2)}\text Q$
2. $\dfrac{1}{4\pi\epsilon_0}\dfrac{(\text R + 2\text r)\text Q}{2(\text R^2 + \text r^2)}$
3. $\dfrac{1}{4\pi\epsilon_0}\dfrac{(\text R + \text r)}{2(\text R^2 + \text r^2)}\text Q$
4. $\dfrac{1}{4\pi\epsilon_0}\dfrac{(2\text R + \text r)}{(\text R^2 + \text r^2)}\text Q$

Answer

$\dfrac{1}{4\pi\epsilon_0}\dfrac{(\text R + \text r)}{(\text R^2 + \text r^2)}\text Q$

Reason —

Given, two concentric shells of radii r and R carrying charges q₁ and q₂ with equal surface charge densities, and q₁ + q₂ = Q.

Since the surface charge densities are equal,

$\dfrac{\text q_1}{4\pi\text r^2} = \dfrac{\text q_2}{4\pi\text R^2}$

Solving these two relations together,

$\text q_1 = \dfrac{\text r^2}{\text R^2 + \text r^2}\text Q, \qquad \text q_2 = \dfrac{\text R^2}{\text R^2 + \text r^2}\text Q$

The potential inside a shell equals that at its surface, so the potential at the common centre is

$\text V = \dfrac{1}{4\pi\epsilon_0}\left(\dfrac{\text q_1}{\text r} + \dfrac{\text q_2}{\text R}\right) \\[1em] = \dfrac{1}{4\pi\epsilon_0}\left(\dfrac{\text r\text Q}{\text R^2 + \text r^2} + \dfrac{\text R\text Q}{\text R^2 + \text r^2}\right) \\[1em] = \dfrac{1}{4\pi\epsilon_0}\cdot\dfrac{(\text R + \text r)}{(\text R^2 + \text r^2)}\text Q$

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

1. 1980 V
2. 660 V
3. 1320 V
4. 1520 V

Answer

1980 V

Reason —

Given, n = 27 and the potential of each small drop is 220 V.

If r and q are the radius and charge of a small drop, and R and 27q those of the bigger drop, then from the conservation of volume,

$\dfrac{4}{3}\pi\text R^3 = 27 \times \dfrac{4}{3}\pi\text r^3 \\[1em] \Rightarrow \quad \text R = 3\text r$

For each small drop,

$220 = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\text r}$

For the bigger drop,

$\text V = \dfrac{1}{4\pi\epsilon_0}\dfrac{27\text q}{3\text r} \\[1em] = 9 \times \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\text r} \\[1em] = 9 \times 220 \\[1em] = 1980\ \text V$

The two thin coaxial rings, each of radius a and having charges +Q and −Q respectively are separated by a distance of s. The potential difference between the centres of the two rings is :

1. $\dfrac{\text Q}{2\pi\epsilon_0}\left[\dfrac{1}{\text a} - \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$
2. $\dfrac{\text Q}{4\pi\epsilon_0}\left[\dfrac{1}{\text a} - \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$
3. $\dfrac{\text Q}{4\pi\epsilon_0}\left[\dfrac{1}{\text a} + \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$
4. $\dfrac{\text Q}{2\pi\epsilon_0}\left[\dfrac{1}{\text a} + \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$

Answer

$\dfrac{\text Q}{2\pi\epsilon_0}\left[\dfrac{1}{\text a} - \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$

Reason —

Given, two coaxial rings A and B, each of radius a, carrying charges + Q and − Q, separated by a distance s.

Every point of a ring is at a distance a from its own centre and at a distance $\sqrt{\text s^2 + \text a^2}$ from the centre of the other ring. Writing $\text K = \dfrac{1}{4\pi\epsilon_0}$,

$\text V_\text A = \dfrac{\text{KQ}}{\text a} - \dfrac{\text{KQ}}{\sqrt{\text a^2 + \text s^2}} \\[1em] \text V_\text B = -\dfrac{\text{KQ}}{\text a} + \dfrac{\text{KQ}}{\sqrt{\text a^2 + \text s^2}}$

Therefore

$\text V_\text A - \text V_\text B = \dfrac{2\text{KQ}}{\text a} - \dfrac{2\text{KQ}}{\sqrt{\text a^2 + \text s^2}} \\[1em] = \dfrac{\text Q}{2\pi\epsilon_0}\left[\dfrac{1}{\text a} - \dfrac{1}{\sqrt{\text s^2 + \text a^2}}\right]$

Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be :

1. 1 : 2
2. 2 : 1
3. 1 : 1
4. 1 : 4

Answer

2 : 1

Reason —

Given, r_A = 5 mm and r_B = 10 mm.

When two conducting spheres are connected by a conducting wire, charge flows from the sphere at the higher potential to the one at the lower potential till both acquire the same potential,

$\text V_\text A = \text V_\text B \\[1em] \Rightarrow \quad \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q_\text A}{\text r_\text A} = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q_\text B}{\text r_\text B} \\[1em] \Rightarrow \quad \dfrac{\text q_\text A}{\text q_\text B} = \dfrac{\text r_\text A}{\text r_\text B} \\[1em] = \dfrac{5}{10} \\[1em] = \dfrac{1}{2}$

The ratio of the electric fields at the surfaces is

$\dfrac{\text E_\text A}{\text E_\text B} = \dfrac{\text K\dfrac{\text q_\text A}{\text r_\text A^2}}{\text K\dfrac{\text q_\text B}{\text r_\text B^2}} \\[1em] = \dfrac{\text q_\text A}{\text q_\text B} \times \dfrac{\text r_\text B^2}{\text r_\text A^2} \\[1em] = \dfrac{1}{2}\left(\dfrac{10}{5}\right)^2 \\[1em] = \dfrac{1}{2} \times 4 \\[1em] = \dfrac{2}{1}$

Hence the ratio is 2 : 1.

Given below are two statements :

Statement I : Electric potential is constant within and at the surface of each conductor.

Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.

In the light of the above statements, choose the most appropriate answer from the options given below.

1. Both statement I and statement II are correct.
2. Both statement I and statement II are incorrect.
3. Statement I is correct but statement II is incorrect.
4. Statement I is incorrect but statement II is correct.

Answer

Both statement I and statement II are correct.

Reason — The electric field is zero at every point inside a conductor, so

$\text E = -\dfrac{\text{dV}}{\text{dr}} \\[1em] = 0$

there is no potential gradient within the conductor and the potential is constant throughout its volume and over its surface. Hence Statement I is correct.

The surface of a conductor is therefore an equipotential surface. Since the electric field is always at right angles to an equipotential surface, the field just outside a charged conductor is perpendicular to the surface at every point. Hence Statement II is also correct.

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

1. 1 : 4
2. 4 : 1
3. 1 : 8
4. 8 : 1

Answer

4 : 1

Reason —

Given, n = 64, r = 0.02 m and q = 5 μC.

From the conservation of volume,

$\text n\left(\dfrac{4}{3}\pi\text r^3\right) = \dfrac{4}{3}\pi\text R^3 \\[1em] \Rightarrow \quad \text R = (64)^{1/3}\text r \\[1em] = 4 \times 0.02 \\[1em] = 0.08\ \text m$

The surface charge densities of the small and the big drop are

$\sigma_\text r = \dfrac{\text q}{4\pi\text r^2}, \qquad \sigma_\text R = \dfrac{\text{nq}}{4\pi\text R^2}$

Therefore

$\dfrac{\sigma_\text R}{\sigma_\text r} = \dfrac{\text{nq}}{4\pi\text R^2} \times \dfrac{4\pi\text r^2}{\text q} \\[1em] = \text n\dfrac{\text r^2}{\text R^2} \\[1em] = 64 \times \dfrac{(0.02)^2}{(0.08)^2} \\[1em] = 64 \times \dfrac{1}{16} \\[1em] = \dfrac{4}{1}$

Hence the ratio is 4 : 1.

If the electric potential at any point (x, y, z) m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3) m will be :

1. 3 V m^-1, directed along positive x-axis
2. 3 V m^-1, directed along negative x-axis
3. 6 V m^-1, directed along positive x-axis
4. 6 V m^-1, directed along negative x-axis

Answer

6 V m^-1, directed along negative x-axis

Reason —

Given, V = 3x² volt.

The potential depends only on x, so the field has only an x-component,

$\vec{\text E} = -\dfrac{\text{dV}}{\text{dx}}\hat{\text i} \\[1em] = -\dfrac{\text d}{\text{dx}}(3\text x^2)\hat{\text i} \\[1em] = -3(2\text x)\hat{\text i} \\[1em] = -6\text x\ \hat{\text i}$

At the point (1, 0, 3), x = 1 m,

$\vec{\text E} = -6\ \hat{\text i}\ \text V\text m^{-1}$

The magnitude is 6 V m^-1 and the negative sign shows that it is directed along the negative x-axis.

Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q, and the remaining one has charge x. The perpendicular from each charge to the nearest hexagon side passes through the centre O of the hexagon and is bisected by the side. Which of the following statement(s) is(are) correct in SI units?

1. When x = q, the magnitude of the electric field at O is zero.
2. When x = −q, the magnitude of the electric field at O is $\dfrac{\text q}{6\pi\epsilon_0\text a^2}$.
3. When x = 2q, the potential at O is $\dfrac{7\text q}{4\sqrt3 \pi\epsilon_0\text a}$.
4. When x = − 3q, the potential at O is $-\dfrac{3\text q}{4\sqrt3 \pi\epsilon_0\text a}$.

Answer

Options 1, 2 and 3 are correct.

Reason —

The distance of the centre O of a regular hexagon of side a from any of its sides (the apothem) is $\dfrac{\sqrt3}{2}\text a$. Since the perpendicular drawn from each charge to the nearest side passes through O and is bisected by that side, the distance of each charge from O is

$\text d = 2 \times \dfrac{\sqrt3}{2}\text a \\[1em] = \sqrt3\ \text a$

Option 1 : When x = q, all the six charges are equal and are symmetrically placed about O. The fields due to diametrically opposite charges cancel in pairs, so the resultant electric field at O is zero. Correct.

Option 2 : When x = − q, we may write the sixth charge as q + (− 2q). The six equal charges q give zero field at O, so only the extra charge − 2q contributes,

$\text E = \dfrac{1}{4\pi\epsilon_0}\dfrac{2\text q}{(\sqrt3 \text a)^2} \\[1em] = \dfrac{1}{4\pi\epsilon_0}\dfrac{2\text q}{3\text a^2} \\[1em] = \dfrac{\text q}{6\pi\epsilon_0\text a^2}$

Correct.

Option 3 : When x = 2q, the total charge is 5q + 2q = 7q, and every charge is at the same distance $\sqrt3 \text a$ from O. Since potential is a scalar,

$\text V_\text O = \dfrac{1}{4\pi\epsilon_0}\dfrac{7\text q}{\sqrt3 \text a} \\[1em] = \dfrac{7\text q}{4\sqrt3 \pi\epsilon_0\text a}$

Correct.

Option 4 : When x = − 3q, the total charge is 5q − 3q = 2q, so

$\text V_\text O = \dfrac{1}{4\pi\epsilon_0}\dfrac{2\text q}{\sqrt3 \text a} \\[1em] = \dfrac{\text q}{2\sqrt3 \pi\epsilon_0\text a}$

which is positive and not equal to the value stated. Not correct.

The angle between the electric lines of force and the equipotential surface is :

1. 180°
2. 0°
3. 45°
4. 90°

Answer

90°

Reason — There is no potential gradient along any direction parallel to an equipotential surface, so the electric field can have no component parallel to the surface,

$\text E_{\text{parallel}} = -\dfrac{\text{dV}}{\text{dr}} \\[1em] = 0$

Hence the electric field, and with it the lines of force, are always at right angles to the equipotential surface.

Two hollow conducting spheres of radii R₁ and R₂ (R₁ >> R₂) have equal charges. The potential would be :

1. dependent on the material property of the sphere
2. more on bigger sphere
3. more on smaller sphere
4. equal on both the spheres

Answer

more on smaller sphere

Reason — The potential of a charged conducting sphere is

$\text V = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\text R} \quad \Rightarrow \quad \text V \propto \dfrac{1}{\text R}$

for a given charge q. Since R₁ >> R₂, we get V₂ >> V₁, that is, the potential is more on the smaller sphere. The potential does not depend on the material of the sphere.

An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at point P due to the electric dipole is $\left(\dfrac{1}{4\pi\epsilon_0} = \text K\right)$ :

1. $\dfrac{8}{5}\text q\text K$
2. $\dfrac{8}{3}\text q\text K$
3. $\dfrac{3}{8}\text q\text K$
4. $\dfrac{5}{8}\text q\text K$

Answer

$\dfrac{3}{8}\text q\text K$

Reason —

From the figure, the two charges − q and + q are 3 cm on either side of the centre O, so the half-length of the dipole is l = 3 cm = 3 × 10^-2 m, and the point P lies on the axis at r = 5 cm = 5 × 10^-2 m from O.

The potential at an axial point of a dipole is

$\text V = \text K\dfrac{\text p}{\text r^2 - \text l^2} \\[1em] = \text K\dfrac{\text q(2\text l)}{\text r^2 - \text l^2}$

Substituting the values,

$= \dfrac{\text{Kq}(6\times10^{-2})}{(5^2 - 3^2)\times10^{-4}} \\[1em] = \dfrac{6\text{Kq}\times10^{-2}}{16\times10^{-4}} \\[1em] = \dfrac{6\text{Kq}\times10^2}{16} \\[1em] = \dfrac{3}{8}\text q\text K \times 10^2\ \text V$

Hence, in units of 10² V, the potential at P is $\dfrac{3}{8}\text q\text K$.

The electric potential at the centre of two concentric half rings of radii R₁ and R₂, having same linear charge density λ is :

1. $\dfrac{2\lambda}{\epsilon_0}$
2. $\dfrac{\lambda}{2\epsilon_0}$
3. $\dfrac{\lambda}{4\epsilon_0}$
4. $\dfrac{\lambda}{\epsilon_0}$

Answer

$\dfrac{\lambda}{2\epsilon_0}$

Reason —

Given, two concentric half rings of radii R₁ and R₂ having the same linear charge density λ.

The length of a half ring of radius R is πR, so the charges on the two half rings are

$\text Q_1 = \lambda(\pi\text R_1), \qquad \text Q_2 = \lambda(\pi\text R_2)$

Every element of a half ring is at the same distance from its centre, so the potential at O is

$\text V = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text Q_1}{\text R_1} + \dfrac{1}{4\pi\epsilon_0}\dfrac{\text Q_2}{\text R_2} \\[1em] = \dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda\pi\text R_1}{\text R_1} + \dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda\pi\text R_2}{\text R_2} \\[1em] = \dfrac{\lambda\pi}{4\pi\epsilon_0}(1 + 1) \\[1em] = \dfrac{\lambda}{2\epsilon_0}$

A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is : $\left(\text{Take}\ \dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{SI units}\right)$

1. 3 × 10⁵
2. 1 × 10⁵
3. 0.5 × 10⁵
4. zero

Answer

zero

Reason —

Given, a thin charged spherical shell of radius 3 cm carrying a charge of 1 μC, with C at the centre and P on the surface.

The electric field is zero at every point inside a charged spherical shell, so there is no potential gradient inside it. Hence the potential at every interior point is the same as that at the surface,

$\text V_\text C = \text V_\text P = \dfrac{1}{4\pi\epsilon_0}\dfrac{\text q}{\text R} \\[1em] \Rightarrow \quad \text V_\text C - \text V_\text P = 0$

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector $\vec{\text P} \space$of magnitude, 4 × 10^-6 C-m, is ± 9 × 10³ V. $\left(\text{Take}\ \dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{SI units}\right)$

Reason (R): $\text V = \pm\dfrac{2\text P}{4\pi\epsilon_0\ \text r^2}$, where r is the distance of any axial point, situated at 2 m from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below :

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, but R is false.
4. A is false, but R is true.

Answer

A is true, but R is false.

Reason —

Explanation

Assertion (A) is correct : The potential at any point on the axis of a short dipole is

$\text V = \pm\dfrac{1}{4\pi\epsilon_0}\dfrac{|\vec{\text P}|}{\text r^2}$

Substituting P = 4 × 10^-6 C-m and r = 2 m,

$\text V = \pm\dfrac{(9\times10^9)(4\times10^{-6})}{(2)^2} \\[1em] = \pm\dfrac{3.6\times10^4}{4} \\[1em] = \pm 9\times10^3\ \text V$

which is exactly the value stated.

Reason (R) is incorrect : The expression given carries an extra factor of 2. The correct relation for the potential at an axial point of a short dipole is $\text V = \pm\dfrac{\text P}{4\pi\epsilon_0\text r^2}$, without the factor 2. The factor 2 belongs to the expression for the electric intensity at an axial point, $\text E = \dfrac{2\text P}{4\pi\epsilon_0\text r^3}$.

Therefore, Assertion is true but Reason is false.

The average translational kinetic energy of N₂ gas molecules at ...... °C becomes equal to the KE of an electron accelerated from rest through a potential difference of 0.1 V.

[Given, K_B = 1.38 × 10^-23 J/K]

Answer

Given,

• Potential difference, V = 0.1 V
• Charge on the electron, e = 1.6 × 10^-19 C
• Boltzmann constant, K_B = 1.38 × 10^-23 J/K

The average translational kinetic energy of a gas molecule is

$\text{KE} = \dfrac{3}{2}\text K_\text B\text T$

and the kinetic energy of the electron accelerated through a potential difference V is eV. Equating the two,

$\dfrac{3}{2}\text K_\text B\text T = \text e\text V$

Substituting the values,

$\dfrac{3}{2}(1.38\times10^{-23})\text T = (1.6\times10^{-19})(0.1) \\[1em] 2.07\times10^{-23}\ \text T = 1.6\times10^{-20} \\[1em] \text T = \dfrac{1.6\times10^{-20}}{2.07\times10^{-23}} \\[1em] = 773\ \text K$

Converting to the Celsius scale,

$\text t = 773 - 273 \\[1em] = 500\ °\text C$

Hence, the required temperature is 500 °C.

27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ...... times that of a smaller drop.

Answer

Given,

• Number of drops, n = 27
• Potential of each small drop, V₁ = 10 V

Let q₁ and r₁ be the charge and radius of a small drop, and Q and R those of the bigger drop. Since the volume remains constant,

$\dfrac{4}{3}\pi\text R^3 = 27 \times \dfrac{4}{3}\pi\text r_1^3 \\[1em] \Rightarrow \quad \text R = 3\text r_1$

Also, the charge is conserved, so Q = 27q₁.

The electrostatic potential energy of a charged spherical drop is

$\text U = \dfrac{3}{5}\dfrac{\text{kq}^2}{\text r}$

For a small drop :

$\text U_\text s = \dfrac{3}{5}\dfrac{\text{kq}_1^2}{\text r_1} \qquad \dots(\text i)$

For the bigger drop :

$\text U_\text B = \dfrac{3}{5}\dfrac{\text k(27\text q_1)^2}{3\text r_1} \\[1em] = \dfrac{3}{5}\dfrac{\text k(27)^2\text q_1^2}{3\text r_1} \qquad \dots(\text{ii})$

Dividing Eq. (ii) by Eq. (i),

$\dfrac{\text U_\text B}{\text U_\text s} = \dfrac{(27)^2}{3} \\[1em] = \dfrac{729}{3} \\[1em] = 243$

Hence, the potential energy of the bigger drop is 243 times that of a smaller drop.

512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ...... V.

Answer

Given,

• Number of drops, n = 512
• Potential of each drop, V = 2 V

Let q and r be the charge and radius of a small drop, and Q and R those of the single big drop. Since the volume remains constant when the drops join,

$512 \times \dfrac{4}{3}\pi\text r^3 = \dfrac{4}{3}\pi\text R^3 \\[1em] \Rightarrow \quad \text R = (512)^{1/3}\text r \\[1em] = 8\text r$

For a small drop,

$\text V = \dfrac{\text{kq}}{\text r} \\[1em] = 2 \qquad \dots(\text i)$

For the big drop, whose charge is Q = 512q,

$\text V_{\text{net}} = \dfrac{\text k(512\text q)}{8\text r} \qquad \dots(\text{ii})$

Dividing Eq. (ii) by Eq. (i),

$\dfrac{\text V_{\text{net}}}{\text V} = \dfrac{512}{8} \\[1em] = 64 \\[1em] \text V_{\text{net}} = 64 \times 2 \\[1em] = 128\ \text V$

Hence, the potential of the single big drop is 128 V.

Two point charges −Q and $+\dfrac{\text Q}{\sqrt3}$ are placed in the xy-plane at the origin (0, 0) and a point (2, 0) respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its centre at (b, 0). All lengths are measured in metres.

The value of R is ............ metre.

Answer

Given,

• Charge − Q at the origin (0, 0)
• Charge $+\dfrac{\text Q}{\sqrt3}$ at the point (2, 0)

Let (x, y) be any point of the zero-potential circle. Since potential is a scalar,

$\text V(\text x, \text y) = \dfrac{1}{4\pi\epsilon_0}\left(-\dfrac{\text Q}{\sqrt{\text x^2 + \text y^2}} + \dfrac{\text Q}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}}\right) \\[1em] = 0 \\[1em] \Rightarrow \quad \dfrac{1}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}} = \dfrac{1}{\sqrt{\text x^2 + \text y^2}}$

Squaring both sides and cross-multiplying,

$\text x^2 + \text y^2 = 3\left[(\text x - 2)^2 + \text y^2\right] \\[1em] \text x^2 + \text y^2 = 3\text x^2 - 12\text x + 12 + 3\text y^2 \\[1em] 2\text x^2 + 2\text y^2 - 12\text x + 12 = 0$

Dividing throughout by 2,

$\text x^2 + \text y^2 - 6\text x + 6 = 0$

Completing the square in x,

$(\text x - 3)^2 + \text y^2 = 9 - 6 \\[1em] = 3 \\[1em] (\text x - 3)^2 + \text y^2 = \left(\sqrt3\right)^2$

Comparing with the standard form (x − h)² + (y − k)² = R²,

$\text R = \sqrt3 \\[1em] = 1.73\ \text m$

Hence, the value of R is 1.73 metre.

Two point charges −Q and $+\dfrac{\text Q}{\sqrt3}$ are placed in the xy-plane at the origin (0, 0) and a point (2, 0) respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its centre at (b, 0). All lengths are measured in metres.

The value of b is ............ metre.

Answer

Given,

• Charge − Q at the origin (0, 0)
• Charge $+\dfrac{\text Q}{\sqrt3}$ at the point (2, 0)
• The zero-potential circle has its centre at (b, 0)

Let (x, y) be any point lying on the circle of zero potential. Its distances from the two charges are

$\text r_1 = \sqrt{\text x^2 + \text y^2}, \qquad \text r_2 = \sqrt{(\text x - 2)^2 + \text y^2}$

Since potential is a scalar quantity, the potentials due to the two charges add algebraically,

$\text V(\text x, \text y) = \dfrac{1}{4\pi\epsilon_0}\left(-\dfrac{\text Q}{\sqrt{\text x^2 + \text y^2}} + \dfrac{\text Q}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}}\right) \\[1em] = 0$

Cancelling $\dfrac{\text Q}{4\pi\epsilon_0}$ and transposing,