Electric Charges & Fields — Competition Zone

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of falls of the proton is :

1. smaller
2. equal
3. 10 times greater
4. 5 times greater

Answer

smaller

Reason —

Given, both the particles fall from rest through the same vertical distance h, the magnitude of the electric field being the same in the two cases.

The electric force on a particle of charge q is F = qE, so its acceleration is

$\text a = \dfrac{\text{qE}}{\text m}$

Starting from rest and covering a distance h,

$\text h = \dfrac{1}{2}\text a\text t^2 \\[1em] \text h = \dfrac{1}{2}\left(\dfrac{\text{qE}}{\text m}\right)\text t^2 \\[1em] \text t = \sqrt{\dfrac{2\text{hm}}{\text{qE}}} \quad \Rightarrow \quad \text t \propto \sqrt{\text m}$

since h, q and E are the same for the electron and for the proton. The mass of the electron is much smaller than that of the proton, so the time of fall of the electron is smaller than that of the proton.

Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :

1. $\dfrac{16\text F}{9}$
2. $\dfrac{4\text F}{3}$
3. F
4. $\dfrac{9\text F}{16}$

Answer

$\dfrac{9\text F}{16}$

Reason —

Given, the charges are + Q and − Q, a distance r apart, and the force between them is

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q^2}{\text r^2}$

When 25% of the charge of A is transferred to B, the charge transferred is $\dfrac{\text Q}{4}$, so the new charges are

$\text Q_\text A = \text Q - \dfrac{\text Q}{4} \\[1em] \text Q_\text A = \dfrac{3\text Q}{4} \\[1em] \text Q_\text B = -\text Q + \dfrac{\text Q}{4} \\[1em] \text Q_\text B = -\dfrac{3\text Q}{4}$

The separation is unchanged, so the new force is

$\text F' = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\left(\dfrac{3\text Q}{4}\right)^2}{\text r^2} \\[1em] \text F' = \dfrac{9}{16} \times \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q^2}{\text r^2} \\[1em] \text F' = \dfrac{9\text F}{16}$

Three charges +Q, q, +Q are placed respectively at distance $0, \dfrac{\text d}{2}$ and d from the origin on the X-axis. If the net force experienced by +Q placed at x = 0 is zero, then value of q is :

1. $\dfrac{+\text Q}{2}$
2. $\dfrac{+\text Q}{4}$
3. $\dfrac{-\text Q}{2}$
4. $\dfrac{-\text Q}{4}$

Answer

$\dfrac{-\text Q}{4}$

Reason —

Given, + Q at x = 0, q at $\text x = \dfrac{\text d}{2}$ and + Q at x = d.

The charge + Q at x = 0 experiences

• a repulsion due to the charge + Q at x = d, at a distance d,

$\text F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q^2}{\text d^2}$

• a force due to the charge q at a distance $\dfrac{\text d}{2}$,

$\text F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\left(\dfrac{\text d}{2}\right)^2} \\[1em] \text F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{4\text{Qq}}{\text d^2}$

For the net force on the charge at x = 0 to be zero, these two must cancel,

$\dfrac{\text Q^2}{\text d^2} + \dfrac{4\text{Qq}}{\text d^2} = 0 \\[1em] \text Q + 4\text q = 0 \\[1em] \text q = -\dfrac{\text Q}{4}$

Two point charges q₁ $\left(\sqrt{10}\ μ\text C\right)$ and q₂ (−25 μC) are placed on the X-axis at x = 1 m and x = 4 m, respectively. The electric field (in V/m) at a point y = 3 m on Y-axis is $\left(\text{take}\ \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N-m}^2\text C^{-2}\right)$

1. $\left(63\hat{\text i} - 27\hat{\text j}\right) \times 10^2$
2. $\left(81\hat{\text i} - 81\hat{\text j}\right) \times 10^2$
3. $\left(-81\hat{\text i} + 81\hat{\text j}\right) \times 10^2$
4. $\left(-63\hat{\text i} + 27\hat{\text j}\right) \times 10^2$

Answer

$\left(63\hat{\text i} - 27\hat{\text j}\right) \times 10^2$

Reason —

Given, $\text q_1 = \sqrt{10}\ \mu\text C$ at (1 m, 0), $\text q_2 = -25\ \mu\text C$ at (4 m, 0), and the point of observation is P (0, 3 m).

Field due to q₁ : The vector from q₁ to P is

$\vec{\text r}_1 = (0 - 1)\hat{\text i} + (3 - 0)\hat{\text j} \\[1em] \vec{\text r}_1 = -\hat{\text i} + 3\hat{\text j} \\[1em] \text r_1 = \sqrt{1 + 9} \\[1em] \text r_1 = \sqrt{10}\ \text m \\[1em] \vec{\text E}_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_1}{\text r_1^{\ 3}}\vec{\text r}_1 \\[1em] \vec{\text E}_1 = \dfrac{\left(9 \times 10^9\right)\left(\sqrt{10} \times 10^{-6}\right)}{10\sqrt{10}}\left(-\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_1 = 900\left(-\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_1 = \left(-900\hat{\text i} + 2700\hat{\text j}\right)\ \text V/\text m$

Field due to q₂ : The vector from q₂ to P is

$\vec{\text r}_2 = (0 - 4)\hat{\text i} + (3 - 0)\hat{\text j} \\[1em] \vec{\text r}_2 = -4\hat{\text i} + 3\hat{\text j} \\[1em] \text r_2 = \sqrt{16 + 9} \\[1em] \text r_2 = 5\ \text m \\[1em] \vec{\text E}_2 = \dfrac{\left(9 \times 10^9\right)\left(-25 \times 10^{-6}\right)}{125}\left(-4\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_2 = -1800\left(-4\hat{\text i} + 3\hat{\text j}\right) \\[1em] = \left(7200\hat{\text i} - 5400\hat{\text j}\right)\ \text V/\text m$

Resultant field :

$\vec{\text E} = \vec{\text E}_1 + \vec{\text E}_2 \\[1em] \vec{\text E} = \left(-900 + 7200\right)\hat{\text i} + \left(2700 - 5400\right)\hat{\text j} \\[1em] \vec{\text E} = \left(6300\hat{\text i} - 2700\hat{\text j}\right) \\[1em] \vec{\text E} = \left(63\hat{\text i} - 27\hat{\text j}\right) \times 10^2\ \text V/\text m$

Four point charges −q, +q, +q and −q are placed on Y-axis at y = −2d, y = −d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at X = D, with D >> d, will behave as :

1. $\text E \propto \dfrac{1}{\text D}$
2. $\text E \propto \dfrac{1}{\text D^3}$
3. $\text E \propto \dfrac{1}{\text D^2}$
4. $\text E \propto \dfrac{1}{\text D^4}$

Answer

$\text E \propto \dfrac{1}{\text D^4}$

Reason —

Given, the charges − q, + q, + q and − q are placed on the Y-axis at y = − 2d, − d, + d and + 2d respectively.

The charges + q at y = + d and − q at y = + 2d form one dipole, and the charges + q at y = − d and − q at y = − 2d form a second dipole of equal moment pointing in the opposite direction. The algebraic sum of all the charges is zero and the net dipole moment of the whole arrangement is also zero.

Such an arrangement of two equal and opposite dipoles is a quadrupole. Just as the field of a dipole falls off one power of distance faster than that of a point-charge, the field of a quadrupole falls off one power faster than that of a dipole,

$\text{point-charge} : \text E \propto \dfrac{1}{\text D^2},$

$\text{dipole} : \text E \propto \dfrac{1}{\text D^3},$

$\text{quadrupole} : \text E \propto \dfrac{1}{\text D^4}$

Hence at a point on the x-axis with D >> d, the field behaves as $\text E \propto \dfrac{1}{\text D^4}$.

The bob of a simple pendulum has mass 2g and a charge of 5.0 μC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take g = 10 m/s²)

1. tan^-1(5.0)
2. tan^-1(2.0)
3. tan^-1(0.5)
4. tan^-1(0.2)

Answer

tan^-1(0.5)

Reason —

Given,

• Mass of the bob, m = 2 g = 2 × 10^-3 kg
• Charge, q = 5.0 μC = 5.0 × 10^-6 C
• Electric field, E = 2000 V/m, horizontal
• g = 10 m/s²

At equilibrium the string makes an angle θ with the vertical. Resolving the tension into horizontal and vertical components,

$\text T\sin\theta = \text{qE}, \qquad \text T\cos\theta = \text{mg}$

Dividing,

$\tan\theta = \dfrac{\text{qE}}{\text{mg}}$

Substituting the values,

$\tan\theta = \dfrac{\left(5.0 \times 10^{-6}\right)(2000)}{\left(2 \times 10^{-3}\right)(10)} \\[1em] \tan\theta = \dfrac{1.0 \times 10^{-2}}{2 \times 10^{-2}} \\[1em] \tan\theta = 0.5 \\[1em] \theta = \tan^{-1}(0.5)$

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then, value of h is :

1. $\dfrac{\text R}{\sqrt 2}$
2. $\text R\sqrt 2$
3. R
4. $\dfrac{\text R}{\sqrt 5}$

Answer

$\dfrac{\text R}{\sqrt 2}$

Reason —

The electric field on the axis of a uniformly charged ring of radius R, at a distance h from its centre, is

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q\text h}{\left(\text R^2 + \text h^2\right)^{3/2}}$

For E to be maximum, $\dfrac{\text{dE}}{\text{dh}} = 0$. Differentiating with respect to h,

$\dfrac{\text{dE}}{\text{dh}} = \dfrac{\text Q}{4\pi\varepsilon_0}\left[\dfrac{\left(\text R^2 + \text h^2\right)^{3/2} - \text h \times \dfrac{3}{2}\left(\text R^2 + \text h^2\right)^{1/2}(2\text h)}{\left(\text R^2 + \text h^2\right)^{3}}\right] \\[1em] = \dfrac{\text Q}{4\pi\varepsilon_0}\left[\dfrac{\left(\text R^2 + \text h^2\right)^{1/2}\left[\left(\text R^2 + \text h^2\right) - 3\text h^2\right]}{\left(\text R^2 + \text h^2\right)^{3}}\right]$

Putting this equal to zero,

$\text R^2 + \text h^2 - 3\text h^2 = 0 \\[1em] \text R^2 = 2\text h^2 \\[1em] \text h = \dfrac{\text R}{\sqrt{2}}$

Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure.

1. $\sqrt{3},\text{ql}\ \dfrac{\hat{\text j} - \hat{\text i}}{\sqrt 2}$
2. $2\text{ql}\ \hat{\text j}$
3. $-\sqrt{3},\text{ql}\ \hat{\text j}$
4. $\left(\text{ql}\right)\dfrac{\hat{\text i} + \hat{\text j}}{\sqrt 2}$

Answer

$-\sqrt{3}\thinspace\text{ql}\ \hat{\text j}$

Reason —

From the figure the charge + q lies at the origin, the second charge + q at (l, 0) and the charge − 2q at the apex $\left(\dfrac{\text l}{2},\ \dfrac{\sqrt{3}\text l}{2}\right)$ of the equilateral triangle of side l.

The dipole moment of a system of charges is $\vec{\text p} = \sum \text q_\text i\vec{\text r}_\text i$. Since the algebraic sum of the charges is q + q − 2q = 0, this is independent of the choice of origin.

x-component :

$\text p_\text x = \text q(0) + \text q(\text l) + (-2\text q)\left(\dfrac{\text l}{2}\right) \\[1em] \text p_\text x = \text{ql} - \text{ql} \\[1em] \text p_\text x = 0$

y-component :

$\text p_\text y = \text q(0) + \text q(0) + (-2\text q)\left(\dfrac{\sqrt{3}\text l}{2}\right) \\[1em] \text p_\text y = -\sqrt{3}\ \text{ql} \\[1em] \vec{\text p} = -\sqrt{3}\ \text{ql}\ \hat{\text j}$

An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is :

1. $\sqrt{\dfrac{\text{qE}}{2\ \text{md}}}$
2. $2\sqrt{\dfrac{\text{qE}}{\text{md}}}$
3. $\sqrt{\dfrac{2\text{qE}}{\text{md}}}$
4. $\sqrt{\dfrac{\text{qE}}{\text{md}}}$

Answer

$\sqrt{\dfrac{2\text{qE}}{\text{md}}}$

Reason —

Given, the two charges q are of mass m each and are separated by a distance d, so the dipole moment is p = qd.

Moment of inertia about the centre : Each charge is at a distance $\dfrac{\text d}{2}$ from the mid-point,

$\text I = 2\text m\left(\dfrac{\text d}{2}\right)^2 \\[1em] \text I = \dfrac{\text m\text d^2}{2}$

Restoring torque : When the dipole is rotated through a small angle θ from its equilibrium orientation,

$\tau = -\text{pE}\sin\theta \approx -\text{pE}\theta \qquad (\text{for small }\theta)$

This is of the form τ = − cθ, so the motion is angular simple harmonic with

$\omega = \sqrt{\dfrac{\text{pE}}{\text I}}$

Substituting p = qd and $\text I = \dfrac{\text m\text d^2}{2}$,

$\omega = \sqrt{\dfrac{\text{qdE}}{\dfrac{\text m\text d^2}{2}}} \\[1em] \omega = \sqrt{\dfrac{2\text{qE}}{\text{md}}}$

A particle of charge q and mass m is subjected to an electric field E = E₀ (1 − ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

1. $\sqrt{\dfrac{3}{\text a}}$
2. $\sqrt{\dfrac{2}{\text a}}$
3. $\sqrt{\dfrac{1}{\text a}}$
4. a

Answer

$\sqrt{\dfrac{3}{\text a}}$

Reason —

Given, $\text E = \text E_0\left(1 - \text{ax}^2\right)$ along the x-direction, and the particle of charge q starts from rest at x = 0.

The particle starts from rest and again comes to rest at some distance d, so its kinetic energy is zero at both positions. By the work-energy theorem the total work done by the electric force over this distance must be zero,

$\text W = \int_0^\text d \text{qE}\ \text{dx} \\[1em] \text W = \text q\text E_0\int_0^\text d\left(1 - \text{ax}^2\right)\text{dx} \\[1em] \text W = 0$

Carrying out the integration,

$\text q\text E_0\left[\text x - \dfrac{\text{ax}^3}{3}\right]_0^\text d = 0 \\[1em] \text d - \dfrac{\text{ad}^3}{3} = 0 \\[1em] \text d\left(1 - \dfrac{\text{ad}^2}{3}\right) = 0$

Since d ≠ 0 (the initial position is excluded),

$\text d^2 = \dfrac{3}{\text a} \\[1em] \text d = \sqrt{\dfrac{3}{\text a}}$

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is :

(m_e ≈ 9 × 10^-31 kg, e = 1.6 × 10^-19 C)

$\left(\text{Take}\ \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{Nm}^2\ \text C^{-2}\right)$

1. 10²⁵ m/s²
2. 10²⁴ m/s²
3. 10²³ m/s²
4. 10²² m/s²

Answer

10²² m/s²

Reason —

Given,

• Separation, r = 1.6 Å = 1.6 × 10^-10 m
• e = 1.6 × 10^-19 C, m_e = 9 × 10^-31 kg

Electrostatic force of attraction :

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text e^2}{\text r^2} \\[1em] \text F = \dfrac{\left(9 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{\left(1.6 \times 10^{-10}\right)^2} \\[1em] \text F = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{2.56 \times 10^{-20}} \\[1em] \text F = 9 \times 10^{-9}\ \text N$

Acceleration of the electron :

$\text a = \dfrac{\text F}{\text m_\text e} \\[1em] \text a = \dfrac{9 \times 10^{-9}}{9 \times 10^{-31}} \\[1em] \text a = 10^{22}\ \text m/\text s^2$

The electric field at a point on the equatorial plane at a distance r from the centre of a dipole having dipole moment $\vec{\text p}$ is given by :

(r >> separation of two charges forming the dipole, ε₀ = permittivity of free space)

1. $\vec{\text E} = -\dfrac{\vec{\text p}}{4\pi\varepsilon_0\text r^3}$
2. $\vec{\text E} = \dfrac{\vec{\text p}}{4\pi\varepsilon_0\text r^3}$
3. $\vec{\text E} = \dfrac{2\vec{\text p}}{4\pi\varepsilon_0\text r^3}$
4. $\vec{\text E} = -\dfrac{\vec{\text p}}{4\pi\varepsilon_0\text r^2}$

Answer

$\vec{\text E} = -\dfrac{\vec{\text p}}{4\pi\varepsilon_0\text r^3}$

Reason —

For a point in the broad-side-on position, that is, on the equatorial plane at a distance r from the centre of the dipole, the components of the two field intensities perpendicular to the dipole axis cancel while the parallel components add up, giving

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}$

Since r >> the separation of the two charges, l² may be neglected in comparison with r²,

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text p}{\text r^3}$

The direction of the field at every point of the equatorial line is antiparallel to the dipole axis, so a minus sign must be put in the vector form,

$\vec{\text E} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{\text p}}{\text r^3}$

Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?

1. $\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\text r^2}$ for all r
2. $\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{qQ}}{\text R^2} \gt \text F \gt 0$ for r < R
3. $\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\text r^2}$ for r > R
4. $\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\text R^2}$ for r < R

Answer

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\text r^2}$ for r > R

Reason —

A uniformly charged spherical shell behaves, for all points outside it, exactly as though its entire charge Q were concentrated at its centre. Hence for r > R the force on a charge q placed at a distance r from the centre is

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text{Qq}}{\text r^2}$

For points inside the shell, that is, for r < R, the contributions of all the elements of the shell cancel exactly, so the field and hence the force is zero,

$\text F = 0 \qquad \text{for r} \lt \text R$

Charges Q₁ and Q₂ are at points A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then Q₁/Q₂ is proportional to :

1. $\dfrac{\text x_1^3}{\text x_2^3}$

2. $\dfrac{\text x_2^2}{\text x_1^2}$

3. $\dfrac{\text x_1}{\text x_2}$

4. $\dfrac{\text x_2}{\text x_1}$

Answer

$\dfrac{\text x_1}{\text x_2}$

Reason —

Given, from the figure Q₁ is at A on the y-axis at a distance x₁ from O, and Q₂ is at B on the x-axis at a distance x₂ from O.

The field at O due to Q₁ is directed along AO, that is, along the y-axis, and the field due to Q₂ is directed along BO, that is, along the x-axis,

$\text E_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q_1}{\text x_1^{\ 2}}, \qquad \text E_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text Q_2}{\text x_2^{\ 2}}$

If the resultant field at O makes an angle α with the x-axis, then

$\tan\alpha = \dfrac{\text E_1}{\text E_2}$

The hypotenuse AB makes an angle β with the x-axis such that

$\tan\beta = \dfrac{0 - \text x_1}{\text x_2 - 0} \\[1em] \tan\beta = -\dfrac{\text x_1}{\text x_2}$

For the resultant to be perpendicular to AB, the product of the two slopes must be − 1,

$\dfrac{\text E_1}{\text E_2} \times \dfrac{\text x_1}{\text x_2} = 1 \\[1em] \dfrac{\text E_1}{\text E_2} = \dfrac{\text x_2}{\text x_1}$

Substituting the field intensities,

$\dfrac{\text Q_1}{\text x_1^{\ 2}} \times \dfrac{\text x_2^{\ 2}}{\text Q_2} = \dfrac{\text x_2}{\text x_1} \\[1em] \dfrac{\text Q_1}{\text Q_2} = \dfrac{\text x_1^{\ 2}}{\text x_2^{\ 2}} \times \dfrac{\text x_2}{\text x_1} \\[1em] \dfrac{\text Q_1}{\text Q_2} = \dfrac{\text x_1}{\text x_2}$

A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass? (Curves are drawn schematically and are not to scale).

Answer

The straight line passing through the origin, that is, option 2.

Reason —

Given, the point mass carries a positive charge q and is released from rest at the edge of the table. The electric field E is uniform and horizontal, along the x-direction, while gravity acts vertically downwards along the y-direction.

Acceleration along the x-direction :

$\text a_\text x = \dfrac{\text{qE}}{\text m} = \text{a constant}$

Acceleration along the y-direction :

$\text a_\text y = \text g = \text{a constant}$

Since the mass starts from rest, after a time t,

$\text x = \dfrac{1}{2}\text a_\text x\text t^2, \qquad \text y = \dfrac{1}{2}\text g\text t^2$

Dividing one by the other, t² cancels,

$\dfrac{\text y}{\text x} = \dfrac{\text g}{\text a_\text x} \\[1em] \dfrac{\text y}{\text x} = \dfrac{\text{mg}}{\text{qE}} \\[1em] \dfrac{\text y}{\text x} = \text{a constant}$

Hence y is directly proportional to x, and the trajectory of the mass is a straight line passing through the origin.

Note: Both accelerations are constant, so the ratio y/x is fixed and the path must be a straight line. The answer key printed with the question gives option 4, which shows a curved path; the correct option is 2.

A dipole is placed in an electric field as shown. In which direction will it move?

1. Towards the right as its potential energy will increase.
2. Towards the left as its potential energy will increase.
3. Towards the right as its potential energy will decrease.
4. Towards the left as its potential energy will decrease.

Answer

Towards the right as its potential energy will decrease.

Reason —

From the figure the lines of force diverge towards the right, so they are crowded on the left and farther apart on the right. Hence the field is non-uniform, being stronger on the left and weaker on the right.

The dipole is placed with + q on the left and − q on the right, so the dipole moment $\vec{\text p}$, which points from − q to + q, is directed towards the left, that is, antiparallel to $\vec{\text E} \space$.

Net force : The charge + q lies in the stronger part of the field and experiences a force qE₁ towards the right, while the charge − q lies in the weaker part and experiences a force qE₂ towards the left. Since E₁ > E₂, the net force is towards the right.

Potential energy : For a dipole, U = − pE cos θ. Here θ = 180°, so U = + pE. As the dipole moves to the right, the field E becomes weaker, and hence U decreases.

Therefore the dipole moves towards the right and its potential energy decreases.

Two identical tennis balls each having mass m and charge q are suspended from a fixed point by threads of length l. What is the equilibrium separation when each thread makes a small angle θ with the vertical?

1. $\text d = \left(\dfrac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{2}}$
2. $\text d = \left(\dfrac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{3}}$
3. $\text d = \left(\dfrac{\text q^2\text l^2}{2\pi\varepsilon_0\text m^2\text g}\right)^{\frac{1}{3}}$
4. $\text d = \left(\dfrac{\text q^2\text l^2}{2\pi\varepsilon_0\text m^2\text g^2}\right)^{\frac{1}{3}}$

Answer

$\text d = \left(\dfrac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{3}}$

Reason —

Given, two balls each of mass m and charge q hang from a common point by threads of length l, and each thread makes a small angle θ with the vertical. Let the equilibrium separation be d.

Forces on either ball : the weight mg vertically downwards, the tension T along the thread, and the electrostatic repulsion

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\text d^2}$

For equilibrium,

$\text T\sin\theta = \text F, \qquad \text T\cos\theta = \text{mg} \\[1em] \tan\theta = \dfrac{\text F}{\text{mg}}$

For a small angle, $\tan\theta \approx \sin\theta = \dfrac{\text d/2}{\text l} = \dfrac{\text d}{2\text l}$,

$\dfrac{\text d}{2\text l} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\text d^2\text{mg}} \\[1em] \text d^3 = \dfrac{2\text l\text q^2}{4\pi\varepsilon_0\text{mg}} \\[1em] \text d^3 = \dfrac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}} \\[1em] \text d = \left(\dfrac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{3}}$

What will be the magnitude of electric field at point O as shown in figure? Each side of the figure is l and perpendicular to each other.

1. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\text l^2}$
2. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\left(2\text l^2\right)}\left(2\sqrt 2 - 1\right)$
3. $\dfrac{\text q}{4\pi\varepsilon_0\left(2\text l\right)^2}$
4. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{2\text q}{2\text l^2}\left(\sqrt 2\right)$

Answer

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\left(2\text l^2\right)}\left(2\sqrt 2 - 1\right)$

Reason —

Taking O as the origin, with the x-axis horizontal and the y-axis vertical, the positions and charges read off from the figure are

• B (−l, 0) : + q, C (0, l) : + 2q, G (l, 0) : + 2q, F (0, −l) : + q — each at a distance l from O
• A (−l, l) : − q, D (l, l) : + q, E (−l, −l) : + 2q, H (l, −l) : − q — each at a distance $\text l\sqrt 2$ from O

Let $\text k = \dfrac{1}{4\pi\varepsilon_0}$ and $\text E_0 = \dfrac{\text{kq}}{\text l^2}$.

Contribution of the four charges at a distance l : The field of a positive charge at O points away from that charge,

$\vec{\text E}_\text B = \text E_0\hat{\text i}, \quad \vec{\text E}_\text C = -2\text E_0\hat{\text j}, \quad \vec{\text E}_\text G = -2\text E_0\hat{\text i}, \quad \vec{\text E}_\text F = \text E_0\hat{\text j}$

Adding these,

$\vec{\text E}_{(1)} = \left(1 - 2\right)\text E_0\hat{\text i} + \left(-2 + 1\right)\text E_0\hat{\text j} \\[1em] \vec{\text E}_{(1)} = -\text E_0\left(\hat{\text i} + \hat{\text j}\right)$

Contribution of the four charges at a distance $\text l\sqrt 2$ : Here the magnitude of the field of a charge Q is $\dfrac{\text{kQ}}{2\text l^2}$, and the unit vectors are along the diagonals,

$\vec{\text E}_\text A = \dfrac{\text E_0}{2}\dfrac{\left(-\hat{\text i} + \hat{\text j}\right)}{\sqrt 2}, \qquad \vec{\text E}_\text D = \dfrac{\text E_0}{2}\dfrac{\left(-\hat{\text i} - \hat{\text j}\right)}{\sqrt 2} \\[1em] \vec{\text E}_\text E = \text E_0\dfrac{\left(\hat{\text i} + \hat{\text j}\right)}{\sqrt 2}, \qquad \vec{\text E}_\text H = \dfrac{\text E_0}{2}\dfrac{\left(\hat{\text i} - \hat{\text j}\right)}{\sqrt 2}$

Adding the x-components inside the bracket, $-\dfrac{1}{2} - \dfrac{1}{2} + 1 + \dfrac{1}{2} = \dfrac{1}{2}$, and similarly the y-components, $\dfrac{1}{2} - \dfrac{1}{2} + 1 - \dfrac{1}{2} = \dfrac{1}{2}$,

$\vec{\text E}_{(2)} = \dfrac{\text E_0}{2\sqrt 2}\left(\hat{\text i} + \hat{\text j}\right)$

Resultant field :

$\vec{\text E} = \vec{\text E}_{(1)} + \vec{\text E}_{(2)} \\[1em] \vec{\text E} = -\text E_0\left(1 - \dfrac{1}{2\sqrt 2}\right)\left(\hat{\text i} + \hat{\text j}\right) \\[1em] |\vec{\text E}| = \text E_0\left(1 - \dfrac{1}{2\sqrt 2}\right)\sqrt 2 \\[1em] |\vec{\text E}| = \text E_0\left(\sqrt 2 - \dfrac{1}{2}\right) \\[1em] |\vec{\text E}| = \text E_0\dfrac{\left(2\sqrt 2 - 1\right)}{2} \\[1em] \text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\left(2\text l^2\right)}\left(2\sqrt 2 - 1\right)$

Note: The answer key printed with the question gives option 4, which works out to $\sqrt 2\ \text E_0$; the value obtained above from the charges and positions shown in the figure is option 2.

A body having specific charge 8 μC/g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be :

1. 0.5 s
2. 1 s
3. 2 s
4. 3 s

Answer

1 s

Reason —

Given,

• Specific charge, $\dfrac{\text q}{\text m} = 8\ \mu\text C/\text g = \dfrac{8 \times 10^{-6}}{10^{-3}} = 8 \times 10^{-3}$ C/kg
• Electric field, E = 100 V/m
• Distance from the wall, s = 10 cm = 0.1 m

Acceleration of the body :

$\text a = \dfrac{\text{qE}}{\text m} \\[1em] \text a = \left(\dfrac{\text q}{\text m}\right)\text E \\[1em] \text a = \left(8 \times 10^{-3}\right)(100) \\[1em] \text a = 0.8\ \text m/\text s^2$

Time to reach the wall : The body starts from rest, so

$\text s = \dfrac{1}{2}\text a\text t^2 \\[1em] 0.1 = \dfrac{1}{2}(0.8)\text t^2 \\[1em] \text t^2 = \dfrac{0.2}{0.8} \\[1em] \text t^2 = 0.25 \\[1em] \text t = 0.5\ \text s$

Time period : The collision with the wall is perfectly elastic, so the body rebounds with the same speed and is then retarded by the same electric force. It therefore takes an equal time of 0.5 s to return to its starting point, where it comes momentarily to rest, and the motion repeats.

$\text T = 2\text t \\[1em] \text T = 2 \times 0.5 \\[1em] \text T = 1\ \text s$

Note: The printed figure labels the field as 10 V/m, but the question stem gives 100 V/m. The stem value is used; it gives the keyed answer, 1 s.

An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200 N/C as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.

$\left[\text{Take, g} = 9.8\ \text{m/s}^2,\ \sin 30° = \dfrac{1}{2},\ \cos 30° = \dfrac{\sqrt 3}{2}\right]$

1. 0.92 s
2. 0.46 s
3. 2.3 s
4. 1.3 s

Answer

1.3 s

Reason —

Given,

• Angle of the incline, θ = 30°; height, h = 1 m
• Mass, m = 1 kg; charge, q = 5 mC = 5 × 10^-3 C
• Horizontal electric field, E = 200 N/C
• Coefficient of friction, μ = 0.2; g = 9.8 m/s²

Length of the incline :

$\text L = \dfrac{\text h}{\sin 30°} \\[1em] \text L = \dfrac{1}{\dfrac{1}{2}} \\[1em] \text L = 2\ \text m$

Electric force :

$\text F = \text{qE} \\[1em] \text F = \left(5 \times 10^{-3}\right)(200) \\[1em] \text F = 1\ \text N, \quad \text{directed horizontally}$

Forces along the incline : The component of the weight down the incline is mg sin θ, while the horizontal electric force has a component qE cos θ directed up the incline, since it opposes the sliding of the body.

Normal reaction : The normal reaction is the sum of the components of the weight and of the electric force perpendicular to the incline,

$\text N = \text{mg}\cos\theta + \text{qE}\sin\theta \\[1em] \text N = (1)(9.8)\left(\dfrac{\sqrt 3}{2}\right) + (1)\left(\dfrac{1}{2}\right) \\[1em] \text N = 8.487 + 0.5 \\[1em] \text N = 8.987\ \text N$

Force of friction :

$\text f = \mu\text N \\[1em] \text f = (0.2)(8.987) \\[1em] \text f = 1.797\ \text N, \quad \text{up the incline}$

Net force down the incline :

$\text F_{\text{net}} = \text{mg}\sin\theta - \text{qE}\cos\theta - \text f \\[1em] = (1)(9.8)\left(\dfrac{1}{2}\right) - (1)\left(\dfrac{\sqrt 3}{2}\right) - 1.797 \\[1em] \text F_{\text{net}} = 4.9 - 0.866 - 1.797 \\[1em] \text F_{\text{net}} = 2.237\ \text N$

Acceleration :

$\text a = \dfrac{\text F_{\text{net}}}{\text m} \\[1em] \text a = 2.237\ \text m/\text s^2$

Time taken : Starting from rest and covering L = 2 m,

$\text L = \dfrac{1}{2}\text a\text t^2 \\[1em] \quad 2 = \dfrac{1}{2}(2.237)\text t^2\\[1em] \text t^2 = \dfrac{4}{2.237} \\[1em] \text t^2 = 1.788 \\[1em] \text t = 1.3\ \text s$

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is $\text a = \dfrac{\sqrt 3}{2}\text L$.

1. $\dfrac{\sqrt 3 \text Q}{4\pi\varepsilon_0\text L^2}$
2. $\dfrac{\text Q}{3\pi\varepsilon_0\text L^2}$
3. $\dfrac{\text Q}{2\sqrt 3 \pi\varepsilon_0\text L^2}$
4. $\dfrac{\text Q}{4\pi\varepsilon_0\text L^2}$

Answer

$\dfrac{\text Q}{2\sqrt 3 \pi\varepsilon_0\text L^2}$

Reason —

Given, a thin wire of length L carries a charge Q, and the point P lies on its perpendicular bisector at a distance $\text a = \dfrac{\sqrt 3}{2}\text L$ from the centre of the rod.

Linear charge density :

$\lambda = \dfrac{\text Q}{\text L}$

For a finite line-charge, the components of the field along the length cancel in pairs about the mid-point, and the field at a point on the perpendicular bisector is

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\lambda}{\text a}\sin\theta$

where θ is the angle subtended at P by half the rod, so that

$\sin\theta = \dfrac{\dfrac{\text L}{2}}{\sqrt{\text a^2 + \dfrac{\text L^2}{4}}}$

Substituting $\text a = \dfrac{\sqrt 3}{2}\text L$,

$\sqrt{\text a^2 + \dfrac{\text L^2}{4}} = \sqrt{\dfrac{3\text L^2}{4} + \dfrac{\text L^2}{4}} \\[1em] \sqrt{\text a^2 + \dfrac{\text L^2}{4}} = \text L \\[1em] \sin\theta = \dfrac{\text L/2}{\text L} \\[1em] \sin\theta = \dfrac{1}{2}$

Hence

$\text E = \dfrac{1}{4\pi\varepsilon_0} \times \dfrac{2\left(\dfrac{\text Q}{\text L}\right)}{\dfrac{\sqrt 3}{2}\text L} \times \dfrac{1}{2} \\[1em] \text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\text Q}{\sqrt 3\ \text L^2} \\[1em] \text E = \dfrac{\text Q}{2\sqrt 3\ \pi\varepsilon_0\text L^2}$

A cube of side a has point charges +Q located at each of its vertices except at the origin, where the charge is −Q. The electric field at the centre of cube is :

1. $\dfrac{-\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$
2. $\dfrac{\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$
3. $\dfrac{-2\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$
4. $\dfrac{2\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$

Answer

$\dfrac{-2\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$

Reason —

Given, a cube of side a with a charge + Q at each of its eight vertices except the origin, where the charge is − Q.

If charges + Q were placed at all the eight vertices, the fields produced at the centre would cancel in pairs, since the charges at the ends of every body diagonal are equidistant from the centre and produce equal and opposite fields. The resultant would then be zero.

The actual arrangement is obtained from this by adding a charge − 2Q at the origin, since (+ Q) + (− 2Q) = − Q. Hence the field at the centre is simply the field due to a single charge − 2Q placed at the origin.

Distance of the centre from the origin : Half the body diagonal of the cube,

$\text r = \dfrac{\sqrt 3}{2}\text a \\[1em] \text r^2 = \dfrac{3\text a^2}{4}$

Magnitude of the field :

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\text Q}{\dfrac{3\text a^2}{4}} \\[1em] \text E = \dfrac{8\text Q}{4\pi\varepsilon_0 \times 3\text a^2} \\[1em] \text E = \dfrac{2\text Q}{3\pi\varepsilon_0\text a^2}$

Direction : The field of a negative charge points towards it, that is, from the centre towards the origin. The unit vector along this direction is $-\dfrac{\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)}{\sqrt 3}$.

$\vec{\text E} = \dfrac{2\text Q}{3\pi\varepsilon_0\text a^2}\times\left(-\dfrac{\hat{\text x} + \hat{\text y} + \hat{\text z}}{\sqrt 3}\right) \\[1em] \vec{\text E} = \dfrac{-2\text Q}{3\sqrt 3\ \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$

Two identical positive charge Q each are fixed at a distance of 2a apart from each other. Another point charge q₀ with mass 'm' is placed at mid-point between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q₀ executes SHM. The time-period of oscillation of charge q₀ will be :

1. $\sqrt{\dfrac{4\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$
2. $\sqrt{\dfrac{\text q_0\text Q}{4\pi^3\varepsilon_0\text{ma}^3}}$
3. $\sqrt{\dfrac{2\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$
4. $\sqrt{\dfrac{8\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$

Answer

$\sqrt{\dfrac{4\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$

Reason —

Given, two charges Q are fixed a distance 2a apart, and a charge q₀ of mass m is placed at the mid-point, that is, at a distance a from each.

Let q₀ be displaced through a small distance x towards the right-hand charge. Its distance from that charge becomes (a − x) and from the charge at A becomes (a + x). The force F₁ due to the nearer charge acts towards the left and the force F₂ due to the charge at A acts towards the right, so, taking the direction of the displacement as positive, the net force on q₀ is

$\text F = \dfrac{\text Q\text q_0}{4\pi\varepsilon_0}\left[\dfrac{1}{(\text a + \text x)^2} - \dfrac{1}{(\text a - \text x)^2}\right] \\[1em] \text F = \dfrac{\text Q\text q_0}{4\pi\varepsilon_0}\left[\dfrac{(\text a - \text x)^2 - (\text a + \text x)^2}{\left(\text a^2 - \text x^2\right)^2}\right] \\[1em] \text F = -\dfrac{\text Q\text q_0}{4\pi\varepsilon_0}\dfrac{4\text{ax}}{\left(\text a^2 - \text x^2\right)^2}$

For a small displacement, x² is negligible compared with a²,

$\text F = -\dfrac{\text Q\text q_0}{4\pi\varepsilon_0}\dfrac{4\text{ax}}{\text a^4} \\[1em] \text F = -\dfrac{\text Q\text q_0\text x}{\pi\varepsilon_0\text a^3}$

The negative sign shows that this force is directed towards the mean position, so the motion is simple harmonic. Comparing with F = − mω²x,

$\text m\omega^2 = \dfrac{\text Q\text q_0}{\pi\varepsilon_0\text a^3} \\[1em] \omega^2 = \dfrac{\text Q\text q_0}{\pi\varepsilon_0\text{ma}^3}$

Time period :

$\text T = \dfrac{2\pi}{\omega} \\[1em] \text T = 2\pi\sqrt{\dfrac{\pi\varepsilon_0\text{ma}^3}{\text Q\text q_0}} \\[1em] \text T = \sqrt{\dfrac{4\pi^2 \times \pi\varepsilon_0\text{ma}^3}{\text q_0\text Q}} \\[1em] \text T = \sqrt{\dfrac{4\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$

Two identical charged particles each having a mass 10 g and charge 2.0 × 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 m s^-2]

1. 12 cm
2. 10 cm
3. 8 cm
4. 5 cm

Answer

12 cm

Reason —

Given,

• Mass of each particle, m = 10 g = 10 × 10^-3 kg
• Charge on each, q = 2.0 × 10^-7 C
• Coefficient of friction, μ = 0.25; g = 10 m s^-2

The particles are in limiting equilibrium, so the electrostatic force of repulsion between them is just balanced by the limiting force of friction,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\text L^2} = \mu\text{mg} \\[1em] \text L^2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\mu\text{mg}}$

Substituting the values,

$\text L^2 = \dfrac{\left(9 \times 10^9\right)\left(2.0 \times 10^{-7}\right)^2}{(0.25)\left(10 \times 10^{-3}\right)(10)} \\[1em] \text L^2 = \dfrac{9 \times 10^9 \times 4 \times 10^{-14}}{2.5 \times 10^{-2}} \\[1em] \text L^2 = \dfrac{3.6 \times 10^{-4}}{2.5 \times 10^{-2}} \\[1em] \text L^2 = 1.44 \times 10^{-2}\ \text m^2 \\[1em] \text L = 0.12\ \text m \\[1em] \text L = 12\ \text{cm}$

A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1 m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation (θ) of the path of the electron as it comes out of the field will be ............... .

1. tan^-1(4)
2. tan^-1(2)
3. $\tan^{-1}\left(\dfrac{1}{3}\right)$
4. tan^-1(3)

Answer

tan^-1(2)

Reason —

Given,

• Electric field, $\text E = \dfrac{8\text m}{\text e}$ V/m
• Length of the plates, L = 1 m
• Speed of entry, u = 2 m/s, parallel to the plates

Acceleration of the electron perpendicular to the plates :

$\text a = \dfrac{\text{eE}}{\text m} \\[1em] \text a = \dfrac{\text e}{\text m}\times\dfrac{8\text m}{\text e} \\[1em] \text a = 8\ \text m/\text s^2$

Time spent inside the field :

$\text t = \dfrac{\text L}{\text u} \\[1em] \text t = \dfrac{1}{2} \\[1em] \text t = 0.5\ \text s$

Velocity components as the electron leaves the field :

$\text v_\text x = \text u \\[1em] \text v_\text x = 2\ \text m/\text s \\[1em] \text v_\text y = \text{at} \\[1em] \text v_\text y = 8 \times 0.5 \\[1em] \text v_\text y = 4\ \text m/\text s$

Angle of deviation :

$\tan\theta = \dfrac{\text v_\text y}{\text v_\text x} \\[1em] \tan\theta = \dfrac{4}{2} \\[1em] \tan\theta = 2 \\[1em] \theta = \tan^{-1}(2)$

A spherically symmetric charge distribution is considered with charge density varying as

$ρ(\text r) = \begin{cases} ρ_0\left(\dfrac{3}{4} - \dfrac{\text r}{\text R}\right) & \text{for r} \le \text R \\[1em] \text{zero} & \text{for r} \gt \text R \end{cases}$

where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :

1. $\dfrac{ρ_0\text r}{4\varepsilon_0}\left(\dfrac{3}{4} - \dfrac{\text r}{\text R}\right)$
2. $\dfrac{ρ_0\text r}{3\varepsilon_0}\left(\dfrac{3}{4} - \dfrac{\text r}{\text R}\right)$
3. $\dfrac{ρ_0\text r}{4\varepsilon_0}\left(1 - \dfrac{\text r}{\text R}\right)$
4. $\dfrac{ρ_0\text r}{5\varepsilon_0}\left(1 - \dfrac{\text r}{\text R}\right)$

Answer

$\dfrac{ρ_0\text r}{4\varepsilon_0}\left(1 - \dfrac{\text r}{\text R}\right)$

Reason —

Given, $ρ(\text r) = ρ_0\left(\dfrac{3}{4} - \dfrac{\text r}{\text R}\right)$ for r ≤ R and zero for r > R.

Charge enclosed within a sphere of radius r : Taking a thin spherical shell of radius r′ and thickness dr′, of volume 4πr′²dr′,

$\text q_{\text{enc}} = \int_0^\text r ρ_0\left(\dfrac{3}{4} - \dfrac{\text r'}{\text R}\right)4\pi\text r'^2\ \text{dr}' \\[1em] = 4\piρ_0\int_0^\text r\left(\dfrac{3\text r'^2}{4} - \dfrac{\text r'^3}{\text R}\right)\text{dr}'$

Carrying out the integration,

$\text q_{\text{enc}} = 4\piρ_0\left[\dfrac{3}{4}\dfrac{\text r'^3}{3} - \dfrac{1}{\text R}\dfrac{\text r'^4}{4}\right]_0^\text r \\[1em] \text q_{\text{enc}} = 4\piρ_0\left[\dfrac{\text r^3}{4} - \dfrac{\text r^4}{4\text R}\right] \\[1em] \text q_{\text{enc}} = \piρ_0\text r^3\left(1 - \dfrac{\text r}{\text R}\right)$

Electric field at P : By the spherical symmetry of the distribution, the charge enclosed behaves as though it were concentrated at the centre,

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_{\text{enc}}}{\text r^2} \\[1em] \text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\piρ_0\text r^3\left(1 - \dfrac{\text r}{\text R}\right)}{\text r^2} \\[1em] \text E = \dfrac{ρ_0\text r}{4\varepsilon_0}\left(1 - \dfrac{\text r}{\text R}\right)$

A long cylindrical volume contains a uniformly distributed charge of density ρ. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

1. $\dfrac{ρ\text{qR}^2}{4\varepsilon_0}$
2. $\dfrac{ρ\text{qR}^2}{2\varepsilon_0}$
3. $\dfrac{\text qρ}{4\varepsilon_0\text R^2}$
4. $\dfrac{4\varepsilon_0\text R^2}{\text qρ}$

Answer

$\dfrac{ρ\text{qR}^2}{4\varepsilon_0}$

Reason —

Given, a long cylinder of radius R carries a uniform volume charge density ρ, and a charge particle q revolves around it in a circular path of radius r (r > R).

Field outside the charged cylinder : For a length L of the cylinder the enclosed charge is ρπR²L, and by the cylindrical symmetry this behaves as a line-charge of linear density

$\lambda = ρ\pi\text R^2$

The field at a distance r from the axis of a line-charge is

$\text E = \dfrac{\lambda}{2\pi\varepsilon_0\text r} \\[1em] \text E = \dfrac{ρ\text R^2}{2\varepsilon_0\text r}$

Condition for circular motion : The electrostatic force supplies the necessary centripetal force,

$\text{qE} = \dfrac{\text m\text v^2}{\text r} \\[1em] \text q\dfrac{ρ\text R^2}{2\varepsilon_0\text r} = \dfrac{\text m\text v^2}{\text r} \\[1em] \text m\text v^2 = \dfrac{\text qρ\text R^2}{2\varepsilon_0}$

Kinetic energy :

$\text K = \dfrac{1}{2}\text m\text v^2 \\[1em] \text K = \dfrac{1}{2}\times\dfrac{\text qρ\text R^2}{2\varepsilon_0} \\[1em] \text K = \dfrac{ρ\text{qR}^2}{4\varepsilon_0}$

Note that the result is independent of the radius r of the circular path.

Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1 m. The ratio of net force on a charged ball to the force between any two charged balls will be :

1. 1 : 1
2. 1 : 4
3. $\sqrt 3$ : 2
4. $\sqrt 3$ : 1

Answer

$\sqrt 3$ : 1

Reason —

Given, three identical balls each of charge 2 C hang from a common point and form an equilateral triangle of side 1 m.

Force between any two balls :

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2)(2)}{(1)^2} \\[1em] \text F = \dfrac{1}{4\pi\varepsilon_0}\times 4$

Net force on one ball : Each ball is repelled by the other two with forces of equal magnitude F, and the angle between these two forces is 60°, since the triangle is equilateral. Hence

$\text F_{\text{net}} = \sqrt{\text F^2 + \text F^2 + 2\text F^2\cos 60°} \\[1em] \text F_{\text{net}} = \sqrt{2\text F^2 + 2\text F^2\left(\dfrac{1}{2}\right)} \\[1em] \text F_{\text{net}} = \sqrt{3\text F^2} \\[1em] \text F_{\text{net}} = \sqrt 3\ \text F$