Electric Charges & Fields — Numericals

Calculate the number of electrons to have a total negative charge of 1 C.

Answer

Given,

• Total negative charge, q = 1 C
• Charge on one electron, e = 1.6 × 10^-19 C

By the quantisation of charge, q = ne, where n is the number of electrons.

$\text n = \dfrac{\text q}{\text e} \\[1em] \text n = \dfrac{1}{1.6 \times 10^{-19}} \\[1em] \text n = 6.25 \times 10^{18}$

Hence, 6.25 × 10¹⁸ electrons are required.

6.25 × 10¹² electrons are removed from a glass rod. (i) What is the magnitude and nature of charge acquired by the glass rod? (ii) What will be the change in the mass of the rod? (mass of electron = 9.0 × 10^-31 kg)

Answer

Given,

• Number of electrons removed, n = 6.25 × 10¹²
• Charge on one electron, e = 1.6 × 10^-19 C
• Mass of one electron, m_e = 9.0 × 10^-31 kg

(i) Charge acquired by the glass rod :

Removal of electrons leaves a deficiency of electrons, so the rod becomes positively charged. By the quantisation of charge,

$\text q = \text{ne} \\[1em] \text q = 6.25 \times 10^{12} \times 1.6 \times 10^{-19} \\[1em] \text q = 1.0 \times 10^{-6}\ \text C \\[1em] \text q = 1.0\ \mu\text C$

Hence, the rod acquires a charge of 1.0 μC, which is positive.

(ii) Change in the mass of the rod :

Since the electrons carry away their mass with them, the mass of the rod decreases by

$\Delta\text m = \text n\text m_\text e \\[1em] \Delta\text m = 6.25 \times 10^{12} \times 9.0 \times 10^{-31} \\[1em] \Delta\text m = 5.625 \times 10^{-18}\ \text{kg}$

Hence, the mass of the rod decreases by 5.625 × 10^-18 kg.

Is the law of conservation of charge is valid in the following reaction?

${}_{20}\text{Ca}^{44} + {}_{1}\text H^{1} \longrightarrow {}_{21}\text{Sc}^{44} + {}_{0}\text n^{1}$

Answer

Yes, the law of conservation of charge is obeyed in the given reaction.

Explanation:
By the law of conservation of charge, the total charge of an isolated system remains constant before and after the reaction.

The charge of a nucleus is Ze, where Z is the atomic number written as the subscript.

Total charge before the reaction :

$\text q_\text i = (20\text e) + (1\text e) \\[1em] \text q_\text i = 21\text e$

Total charge after the reaction :

$\text q_\text f = (21\text e) + (0) \\[1em] \text q_\text f = 21\text e$

Since q_i = q_f, the net charge is the same before and after the reaction.

Calculate the Coulombian force between a proton and an electron separated by 0.8 × 10^-15 m.

Answer

Given,

• Charge on a proton, q₁ = + 1.6 × 10^-19 C
• Charge on an electron, q₂ = − 1.6 × 10^-19 C
• Separation, r = 0.8 × 10^-15 m

By Coulomb's law, the magnitude of the force is

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|\text q_1\text q_2|}{\text r^2}$

Substituting the values,

$\text F = \dfrac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)\left(1.6 \times 10^{-19}\right)}{\left(0.8 \times 10^{-15}\right)^2} \\[1em] \text F = \dfrac{9.0 \times 10^9 \times 2.56 \times 10^{-38}}{6.4 \times 10^{-31}} \\[1em] \text F = \dfrac{2.304 \times 10^{-28}}{6.4 \times 10^{-31}} \\[1em] \text F = 360\ \text N$

Hence, the Coulombian force is 360 N and it is attractive, since the two charges are of opposite signs.

How far apart the two electrons be, if the force between them equals the weight of an electron? What in case of protons?

Answer

Given,

• Charge on each particle, e = 1.6 × 10^-19 C
• Mass of electron, m_e = 9.0 × 10^-31 kg
• Mass of proton, m_p = 1.7 × 10^-27 kg
• g = 9.8 m s^-2

The Coulomb force between the two particles is to be equal to the weight of one particle,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text e^2}{\text r^2} = \text{mg} \\[1em] \text r = \sqrt{\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text e^2}{\text{mg}}}$

For two electrons :

$\text r^2 = \dfrac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{9.0 \times 10^{-31} \times 9.8} \\[1em] \text r^2 = \dfrac{2.304 \times 10^{-28}}{8.82 \times 10^{-30}} \\[1em] \text r^2 = 26.12 \\[1em] \text r = 5.1\ \text m$

For two protons :

$\text r^2 = \dfrac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{1.7 \times 10^{-27} \times 9.8} \\[1em] \text r^2 = \dfrac{2.304 \times 10^{-28}}{1.666 \times 10^{-26}} \\[1em] \text r^2 = 1.383 \times 10^{-2} \\[1em] \text r = 0.117\ \text m$

Hence, the separation is 5.1 m for two electrons and 0.117 m for two protons.

Two positive charges distant 0.1 m apart, repel each other with a force of 18 N. If the sum of the charges be 9 micro-coulomb (μC), then calculate their separate values.

Answer

Given,

• Separation, r = 0.1 m
• Force of repulsion, F = 18 N
• Sum of the charges, q₁ + q₂ = 9 μC

By Coulomb's law,

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_1\text q_2}{\text r^2} \\[1em] 18 = \dfrac{\left(9.0 \times 10^9\right)\text q_1\text q_2}{(0.1)^2} \\[1em] \text q_1\text q_2 = \dfrac{18 \times 0.01}{9.0 \times 10^9} \\[1em] \text q_1\text q_2 = 2 \times 10^{-11}\ \text C^2 \\[1em] \text q_1\text q_2 = 20\ (\mu\text C)^2$

So the two charges satisfy

$\text q_1 + \text q_2 = 9\ \mu\text C, \qquad \text q_1\text q_2 = 20\ (\mu\text C)^2$

They are therefore the roots of the quadratic equation

$\text q^2 - (9\ \mu\text C)\text q + 20\ (\mu\text C)^2 = 0 \\[1em] \text q^2 - (5\ \mu\text C)\text q - (4\ \mu\text C)\text q + 20\ (\mu\text C)^2 = 0 \\[1em] (\text q - 5\ \mu\text C)(\text q - 4\ \mu\text C) = 0 \\[1em] \text q = 5\ \mu\text C \quad \text{or} \quad \text q = 4\ \mu\text C$

Hence, the two charges are 5 μC and 4 μC.

The distance between two negatively-charged dust particles is 1 mm and they repel each other with a force of 4.0 × 10^-5 N. If the charge on one particle is 4 times that of the other, then find out the number of electrons charging them separately. The charge on electron is $\left(\dfrac{5}{3}\right)$ × 10^-19 C.

Answer

Given,

• Separation, r = 1 mm = 1 × 10^-3 m
• Force of repulsion, F = 4.0 × 10^-5 N
• If the charge on one particle is q, that on the other is 4q
• Charge on an electron, $\text e = \dfrac{5}{3} \times 10^{-19}$ C

By Coulomb's law,

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q \times 4\text q}{\text r^2} \\[1em] 4.0 \times 10^{-5} = \dfrac{\left(9.0 \times 10^9\right)\left(4\text q^2\right)}{\left(1 \times 10^{-3}\right)^2} \\[1em] 4\text q^2 = \dfrac{4.0 \times 10^{-5} \times 10^{-6}}{9.0 \times 10^9} \\[1em] 4\text q^2 = \dfrac{4.0 \times 10^{-11}}{9.0 \times 10^9} \\[1em] \text q^2 = 1.111 \times 10^{-21} \\[1em] \text q = 3.333 \times 10^{-11}\ \text C$

The other charge is 4q = 1.333 × 10^-10 C.

Number of electrons on the first particle :

$\text n_1 = \dfrac{\text q}{\text e} \\[1em] \text n_1 = \dfrac{3.333 \times 10^{-11}}{\dfrac{5}{3} \times 10^{-19}} \\[1em] \text n_1 = 2 \times 10^{8}$

Number of electrons on the second particle :

$\text n_2 = 4\text n_1 \\[1em] \text n_2 = 8 \times 10^{8}$

Hence, the two particles carry 2 × 10⁸ and 8 × 10⁸ electrons respectively.

Two point-charges of +2 μC and +6 μC repel each other with a force of 12 N. If each is given an additional charge of −4 μC, then what will be the new force?

Answer

Given,

• Initial charges, q₁ = + 2 μC and q₂ = + 6 μC
• Initial force of repulsion, F = 12 N
• Additional charge given to each = − 4 μC

New charges :

$\text q_1' = 2\ \mu\text C - 4\ \mu\text C \\[1em] \text q_1' = -2\ \mu\text C \\[1em] \text q_2' = 6\ \mu\text C - 4\ \mu\text C \\[1em] \text q_2' = +2\ \mu\text C$

Since the separation r remains unchanged, by Coulomb's law the magnitude of the force $|\text F| \propto |\text q_1\text q_2|$,

$\dfrac{|\text F'|}{|\text F|} = \dfrac{|\text q_1'\text q_2'|}{|\text q_1\text q_2|} \\[1em] \dfrac{|\text F'|}{|\text F|} = \dfrac{2 \times 2}{2 \times 6} \\[1em] \dfrac{|\text F'|}{|\text F|} = \dfrac{1}{3} \\[1em] |\text F'| = \dfrac{12}{3} \\[1em] |\text F'| = 4\ \text N$

Hence, the new force is 4 N, and it is a force of attraction since the charges are now of opposite signs.

Two identical metallic balls A and B have charges +40 and −10 μC respectively. The distance between them is 2.0 m. What is the magnitude and type of force acting between them? They are touched to each other and again separated by a distance of 2.0 m from each other. Calculate the new force between them. What will be the force if one ball is connected to earth?

Answer

Given,

• Charge on ball A, q_A = + 40 μC = 40 × 10^-6 C
• Charge on ball B, q_B = − 10 μC = − 10 × 10^-6 C
• Separation, r = 2.0 m

Initial force :

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|\text q_\text A\text q_\text B|}{\text r^2} \\[1em] \text F = \dfrac{\left(9.0 \times 10^9\right)\left(40 \times 10^{-6}\right)\left(10 \times 10^{-6}\right)}{(2.0)^2} \\[1em] \text F = \dfrac{9.0 \times 10^9 \times 4.0 \times 10^{-10}}{4} \\[1em] \text F = \dfrac{3.6}{4} \\[1em] \text F = 0.90\ \text N$

Since the charges are of opposite signs, this is a force of attraction.

After the balls are touched : The balls are identical, so the charge is shared equally,

$\text q' = \dfrac{\text q_\text A + \text q_\text B}{2} \\[1em] \text q' = \dfrac{40 - 10}{2} \\[1em] \text q' = +15\ \mu\text C \\[1em] \text F' = \dfrac{\left(9.0 \times 10^9\right)\left(15 \times 10^{-6}\right)^2}{(2.0)^2} \\[1em] \text F' = \dfrac{9.0 \times 10^9 \times 2.25 \times 10^{-10}}{4} \\[1em] \text F' = \dfrac{2.025}{4} \\[1em] \text F' = 0.51\ \text N$

Both the balls now carry like charges, so this is a force of repulsion.

If one ball is connected to earth : Its charge flows to the earth and it becomes uncharged. With one of the charges reduced to zero, the force between the balls is zero.

Hence, the forces are 0.90 N (attraction), 0.51 N (repulsion) and zero respectively.

Two similarly and equally charged identical metal spheres A and B repel each other with a force of 2.0 × 10^-5 N. A third identical, uncharged sphere C is touched with A and then placed at the mid-point between A and B. What is net electric force on C?

Answer

Given,

• Spheres A and B carry equal charges, each q, separated by a distance r
• Force of repulsion between A and B, F = 2.0 × 10^-5 N
• Sphere C is identical and initially uncharged

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\text r^2} \\[1em] \text F = 2.0 \times 10^{-5}\ \text N \qquad \dots(\text i)$

Charge after C touches A : Since A and C are identical, the charge q is shared equally,

$\text q_\text A = \text q_\text C = \dfrac{\text q}{2}$

The charge on B remains q. The sphere C is now placed at the mid-point, so it is at a distance $\dfrac{\text r}{2}$ from both A and B.

Force on C due to A :

$\text F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\left(\dfrac{\text q}{2}\right)\left(\dfrac{\text q}{2}\right)}{\left(\dfrac{\text r}{2}\right)^2} \\[1em] \text F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q^2}{\text r^2} \\[1em] \text F_1 = \text F \quad \text{(directed from A to B)}$

Force on C due to B :

$\text F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\left(\dfrac{\text q}{2}\right)(\text q)}{\left(\dfrac{\text r}{2}\right)^2} \\[1em] \text F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\text q^2}{\text r^2} \\[1em] \text F_2 = 2\text F \quad \text{(directed from B to A)}$

Net force on C : The two forces are oppositely directed,

$\text F_{\text{net}} = 2\text F - \text F \\[1em] \text F_{\text{net}} = \text F \\[1em] \text F_{\text{net}} = 2.0 \times 10^{-5}\ \text N$

Hence, the net electric force on C is 2.0 × 10^-5 N, directed towards A.

A particle having a charge of 2.0 × 10^-4 C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum of mass 100 g. What charge should the bob be given so that the string becomes loose?

Answer

Given,

• Charge placed below the bob, Q = 2.0 × 10^-4 C
• Separation, r = 10 cm = 0.10 m
• Mass of the bob, m = 100 g = 0.1 kg
• g = 9.8 m s^-2

The string becomes loose when its tension falls to zero. This happens when the upward electrostatic force of repulsion on the bob just balances its weight,

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

Substituting the values,

$\dfrac{\left(9.0 \times 10^9\right)\left(2.0 \times 10^{-4}\right)\text q}{(0.10)^2} = 0.1 \times 9.8 \\[1em] \dfrac{1.8 \times 10^6 \times \text q}{0.01} = 0.98 \\[1em] 1.8 \times 10^8 \times \text q = 0.98 \\[1em] \text q = \dfrac{0.98}{1.8 \times 10^8} \\[1em] \text q = 5.4 \times 10^{-9}\ \text C$

Hence, the bob should be given a charge of 5.4 × 10^-9 C, of the same sign as Q so that the force is repulsive.

Ten positively charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, ....., 100 cm. The first particle has a charge 1.0 × 10^-8 C, the second 8 × 10^-8 C, the third 27 × 10^-8 C and so on. The tenth particle has charge 1000 × 10^-8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer

Given,

• The n^th particle is at x_n = 10n cm = 0.1n m, with n = 1, 2, 3, ..., 10
• The charges are 1 × 10^-8 C, 8 × 10^-8 C, 27 × 10^-8 C, ..., 1000 × 10^-8 C, that is, q_n = n³ × 10^-8 C
• Test charge at the origin, q₀ = 1 C

All the charges are positive and lie on the same side of the origin, so by the principle of superposition all the forces on the charge at the origin are in the same direction and simply add up.

Force due to the n^th particle :

$\text F_\text n = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_\text n\text q_0}{\text x_\text n^{\ 2}} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\dfrac{\text n^3 \times 10^{-8}}{(0.1\text n)^2} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\dfrac{\text n^3 \times 10^{-8}}{0.01\text n^2} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\left(10^{-6}\right)\text n \\[1em] \text F_\text n = 9 \times 10^3\ \text n\ \text N$

Total force :

$\text F = \sum_{\text n = 1}^{10}\text F_\text n \\[1em] \text F = 9 \times 10^3 \sum_{\text n=1}^{10}\text n \\[1em] \text F = 9 \times 10^3 \times \dfrac{10 \times 11}{2} \\[1em] \text F = 9 \times 10^3 \times 55 \\[1em] \text F = 4.95 \times 10^5\ \text N$

Hence, the magnitude of the electric force on the 1 C charge at the origin is 4.95 × 10⁵ N.

Two charges +e and +e are static at distance 20 cm apart. Space of 8 cm width is filled with dielectric K = 4, in rest of the part there is vacuum. Find the force between the two given charges.

Answer

Given,

• Each charge = + e = 1.6 × 10^-19 C
• Separation, r = 20 cm = 0.20 m
• Thickness of the dielectric slab, t = 8 cm = 0.08 m
• Dielectric constant, K = 4

When the space between the charges is partly filled with a dielectric slab of thickness t and dielectric constant K, the slab is equivalent to an air distance $\text t\sqrt{\text K}$. Hence the effective separation in air is

$\text r_{\text{eff}} = (\text r - \text t) + \text t\sqrt{\text K}$

Substituting the values,

$\text r_{\text{eff}} = (0.20 - 0.08) + 0.08 \times \sqrt{4} \\[1em] \text r_{\text{eff}} = 0.12 + 0.16 \\[1em] \text r_{\text{eff}} = 0.28\ \text m$

The force between the charges is therefore

$\text F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text e^2}{\text r_{\text{eff}}^{\ 2}} \\[1em] \text F = \dfrac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{(0.28)^2} \\[1em] \text F = \dfrac{9.0 \times 10^9 \times 2.56 \times 10^{-38}}{0.0784} \\[1em] \text F = \dfrac{2.304 \times 10^{-28}}{0.0784} \\[1em] \text F = 2.94 \times 10^{-27}\ \text N$

Hence, the force between the two charges is 2.94 × 10^-27 N.

ABC is an equilateral triangle of side 10 m and D is the mid-point of BC. Charges of +100, −100 and +75 μC are placed at B, C and D respectively. Find the force on a +1 μC charge placed at A.

Answer

Given,

• ABC is an equilateral triangle of side 10 m, D is the mid-point of BC
• q_B = + 100 μC, q_C = − 100 μC, q_D = + 75 μC
• Test charge at A, q = + 1 μC

Distances : AB = AC = 10 m, and AD is the altitude of the equilateral triangle,

$\text{AD} = \dfrac{\sqrt{3}}{2} \times 10 \\[1em] \text{AD} = 5\sqrt{3}\ \text m \\[1em] (\text{AD})^2 = 75\ \text m^2$

Force due to the charge at B : It is a repulsion, directed along BA,

$\text F_\text B = \dfrac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(1 \times 10^{-6}\right)}{(10)^2} \\[1em] \text F_\text B = \dfrac{9 \times 10^{-1}}{100} \\[1em] \text F_\text B = 9 \times 10^{-3}\ \text N$

Force due to the charge at C : It is an attraction of the same magnitude, directed along AC,

$\text F_\text C = 9 \times 10^{-3}\ \text N$

Resultant of F_B and F_C : These two forces are equal in magnitude and are inclined at 120° to each other, so their resultant is

$\text F_1 = 2\text F_\text B\cos 60° \\[1em] \text F_1 = 2 \times 9 \times 10^{-3} \times \dfrac{1}{2} \\[1em] \text F_1 = 9 \times 10^{-3}\ \text N$

directed parallel to BC, from B towards C.

Force due to the charge at D : It is a repulsion, directed along DA, that is, perpendicular to BC,

$\text F_\text D = \dfrac{\left(9 \times 10^9\right)\left(75 \times 10^{-6}\right)\left(1 \times 10^{-6}\right)}{75} \\[1em] \text F_\text D = \dfrac{6.75 \times 10^{-1}}{75} \\[1em] \text F_\text D = 9 \times 10^{-3}\ \text N$

Resultant force at A : F₁ and F_D are equal in magnitude and mutually perpendicular,

$\text F = \sqrt{\text F_1^{\ 2} + \text F_\text D^{\ 2}} \\[1em] \text F = \sqrt{2}\times 9 \times 10^{-3} \\[1em] \text F = 9\sqrt{2} \times 10^{-3}\ \text N$

and it makes an angle

$\tan\theta = \dfrac{\text F_\text D}{\text F_1} \\[1em] \tan\theta = 1 \\[1em] \theta = 45°$

Hence, the force on the + 1 μC charge at A is $9\sqrt{2} \times 10^{-3}$ N, acting at 45° to a line parallel to BC.

Two point-charges of 1.0 μC and −0.25 μC are placed in air at a distance of 0.40 m from each other. Find out at which point on the line joining the two charges should a third charge be placed so that no force acts upon it?

Answer

Given,

• q₁ = 1.0 μC, q₂ = − 0.25 μC
• Separation, r = 0.40 m

Since the two charges are of opposite signs, the point of zero force cannot lie between them. It must lie outside the two charges, on the side of the smaller charge q₂.

Let the third charge be placed at a distance x from q₂, so that its distance from q₁ is (0.40 + x).

For no force to act on the third charge, the two forces on it must be equal in magnitude,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{|\text q_1|}{(0.40 + \text x)^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|\text q_2|}{\text x^2} \\[1em] \dfrac{(0.40 + \text x)^2}{\text x^2} = \dfrac{|\text q_1|}{|\text q_2|} \\[1em] \dfrac{(0.40 + \text x)^2}{\text x^2} = \dfrac{1.0}{0.25} \\[1em] \dfrac{(0.40 + \text x)^2}{\text x^2} = 4$

Taking the square root,

$\dfrac{0.40 + \text x}{\text x} = 2 \\[1em] 0.40 + \text x = 2\text x \\[1em] \text x = 0.40\ \text m$

Hence, the third charge should be placed outside the second charge, at a distance of 0.40 m from the − 0.25 μC charge (that is, 0.80 m from the 1.0 μC charge).

Two point-charges +9e and +e are placed at a distance of 16 cm from each other. At what point between these charges should a third charge q be placed so that it remains in equilibrium?

Answer

Given,

• q₁ = + 9e, q₂ = + e
• Separation, r = 16 cm

Both the charges are positive, so the third charge q can remain in equilibrium only at a point between them, where the two forces on it are oppositely directed.

Let q be placed at a distance x from the charge + 9e, so that its distance from + e is (16 − x) cm.

For the equilibrium of q,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{(9\text e)\text q}{\text x^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(\text e)\text q}{(16 - \text x)^2} \\[1em] \dfrac{9}{\text x^2} = \dfrac{1}{(16 - \text x)^2}$

Taking the square root of both sides,

$\dfrac{3}{\text x} = \dfrac{1}{16 - \text x} \\[1em] 3(16 - \text x) = \text x \\[1em] 48 - 3\text x = \text x \\[1em] 4\text x = 48 \\[1em] \text x = 12\ \text{cm}$

Hence, the third charge should be placed at a distance of 12 cm from the charge + 9e (and 4 cm from the charge + e).

Three point-charges q₁, q₂ and q₃ are in line at equal distances. q₂ and q₃ are opposite in sign. Find the magnitude and sign of q₁, if the net force on q₃ is zero.

Answer

Given,

• q₁, q₂ and q₃ lie in a line at equal distances, say d apart
• q₂ and q₃ are opposite in sign
• Net force on q₃ is zero

The distance of q₃ from q₂ is d and from q₁ is 2d.

For the net force on q₃ to be zero, the forces due to q₁ and q₂ must be equal in magnitude and opposite in direction,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_1\text q_3}{(2\text d)^2} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q_2\text q_3}{\text d^2} = 0 \\[1em] \dfrac{\text q_1}{4} + \text q_2 = 0 \\[1em] \text q_1 = -4\text q_2$

Hence the magnitude of q₁ is four times that of q₂, and its sign is opposite to that of q₂. Since q₂ and q₃ are opposite in sign, q₁ has the same sign as q₃.

Hence, q₁ has a magnitude four times that of q₂ and its sign is the same as that of q₃.

Two free particles with charges +q and +4q are at a distance L apart. A third charge is placed so that the entire system is in equilibrium. (a) Find the location, magnitude and nature of the third charge. (b) Show that the equilibrium is unstable.

Answer

Given,

• Charges + q and + 4q, a distance L apart

(a) Location, magnitude and nature of the third charge :

The third charge Q must lie between the two charges, since only there can the two forces on it be oppositely directed. Let it be at a distance x from + q, so that its distance from + 4q is (L − x).

For the equilibrium of Q,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q\text Q}{\text x^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(4\text q)\text Q}{(\text L - \text x)^2} \\[1em] \dfrac{1}{\text x^2} = \dfrac{4}{(\text L - \text x)^2} \\[1em] \dfrac{\text L - \text x}{\text x} = 2 \\[1em] \text L = 3\text x \\[1em] \text x = \dfrac{\text L}{3}$

Thus Q is at a distance $\dfrac{\text L}{3}$ from + q and $\dfrac{2\text L}{3}$ from + 4q.

For the entire system to be in equilibrium, the charge + q must also be in equilibrium. The forces on it are the repulsion due to + 4q at a distance L and the force due to Q at a distance $\dfrac{\text L}{3}$,

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q(4\text q)}{\text L^2} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q\text Q}{\left(\dfrac{\text L}{3}\right)^2} = 0 \\[1em] \dfrac{4\text q}{\text L^2} + \dfrac{9\text Q}{\text L^2} = 0 \\[1em] \text Q = -\dfrac{4\text q}{9}$

Hence, the third charge is $-\dfrac{4\text q}{9}$, that is, it is negative, and it is placed at a distance $\dfrac{\text L}{3}$ from + q and $\dfrac{2\text L}{3}$ from + 4q.

(b) The equilibrium is unstable :

Suppose the negative charge Q is displaced slightly along the line towards + q. The distance from + q decreases and the distance from + 4q increases, so the attraction due to + q increases while that due to + 4q decreases. The net force on Q is then towards + q, that is, in the direction of the displacement itself, and Q moves still farther from its equilibrium position.

Since the net force does not bring the charge back to its original position, the equilibrium is unstable.

At each of the four corners of a square of side a, a charge +q is placed freely. What charge should be placed at the centre of the square so that the whole system be in equilibrium?

Answer

Given,

• Four charges +q at the corners of a square of side a
• A charge Q is to be placed at the centre

Consider the charge +q at the lower-left corner. The distances of the other charges from it are

• a for the two adjacent corners,
• $\text a\sqrt{2}$ for the opposite corner,
• $\dfrac{\text a}{\sqrt{2}}$ for the centre of the square.

Forces due to the other three corner charges : Let $\text k = \dfrac{1}{4\pi\varepsilon_0}$.

The two adjacent charges exert repulsions F₁ and F₂, each of magnitude $\dfrac{\text{kq}^2}{\text a^2}$, and their components along the diagonal are each $\dfrac{\text{kq}^2}{\text a^2}\cos 45^\circ$. The charge at the opposite corner exerts a repulsion F₃ = $\dfrac{\text{kq}^2}{2\text a^2}$ directly along the diagonal.

Hence the resultant outward force F along the diagonal is

$\text F = 2 \times \dfrac{\text{kq}^2}{\text a^2} \times \dfrac{1}{\sqrt{2}} + \dfrac{\text{kq}^2}{2\text a^2} \\[1em] \text F = \dfrac{\text{kq}^2}{\text a^2}\left(\sqrt{2} + \dfrac{1}{2}\right)$

Force due to the charge at the centre : It acts along the diagonal, of magnitude

$\text F_\text Q = \dfrac{\text{kq}|\text Q|}{\left(\dfrac{\text a}{\sqrt{2}}\right)^2} \\[1em] \text F_\text Q = \dfrac{2\text{kq}|\text Q|}{\text a^2}$

Condition of equilibrium : F_Q must be directed inwards and must balance F,

$\dfrac{2\text{kq}|\text Q|}{\text a^2} = \dfrac{\text{kq}^2}{\text a^2}\left(\sqrt{2} + \dfrac{1}{2}\right) \\[1em] |\text Q| = \dfrac{\text q}{2}\left(\sqrt{2} + \dfrac{1}{2}\right) \\[1em] |\text Q| = \dfrac{\text q\left(1 + 2\sqrt{2}\right)}{4}$

Since the force due to Q has to be an attraction, Q must be negative.

$\text Q = -\dfrac{\text q\left(1 + 2\sqrt{2}\right)}{4}$

The charge Q itself is in equilibrium for any value, as the four equal charges at the corners pull it equally from all sides.

Hence, a charge of $-\dfrac{\text q\left(1 + 2\sqrt{2}\right)}{4}$ should be placed at the centre of the square.

Calculate the intensity of the electric field due to a helium nucleus at a distance of 1 Å from the nucleus.

Hint : The helium nucleus has a positive charge equal to that on an α-particle.

Answer

Given,

• The helium nucleus has a positive charge equal to that on an α-particle, q = + 2e = 3.2 × 10^-19 C
• Distance, r = 1 Å = 1 × 10^-10 m

The intensity of the electric field due to a point-charge is

$\text E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\text r^2}$

Substituting the values,

$\text E = \dfrac{\left(9.0 \times 10^9\right)\left(3.2 \times 10^{-19}\right)}{\left(1 \times 10^{-10}\right)^2} \\[1em] \text E = \dfrac{2.88 \times 10^{-9}}{1 \times 10^{-20}} \\[1em] \text E = 2.88 \times 10^{11}\ \text N\ \text C^{-1}$

Hence, the intensity of the electric field is 2.88 × 10¹¹ N C^-1, directed radially outwards from the nucleus.

A charge of 2 μC is placed at point (1, −2, 1) metre. Find the electric field intensity in vector form at point P(1, 2, 4) meter.

Answer

Given,

• Charge, q = 2 μC = 2 × 10^-6 C, placed at (1, −2, 1) m
• Point of observation, P (1, 2, 4) m

Position vector of P relative to the charge :

$\vec{\text r} = (1 - 1)\hat{\text i} + \left(2 - (-2)\right)\hat{\text j} + (4 - 1)\hat{\text k} \\[1em] \vec{\text r} = 4\hat{\text j} + 3\hat{\text k} \\[1em] \text r = \sqrt{0^2 + 4^2 + 3^2} \\[1em] \text r = 5\ \text m$

Electric field intensity in vector form :

$\vec{\text E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\text r^2}\hat{\text r} \\[1em] \vec{\text E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text q}{\text r^3}\vec{\text r}$

Substituting the values,

$\vec{\text E} = \dfrac{\left(9 \times 10^9\right)\left(2 \times 10^{-6}\right)}{(5)^3}\left(4\hat{\text j} + 3\hat{\text k}\right) \\[1em] \vec{\text E} = \dfrac{1.8 \times 10^4}{125}\left(4\hat{\text j} + 3\hat{\text k}\right) \\[1em] \vec{\text E} = 144\left(4\hat{\text j} + 3\hat{\text k}\right)\ \text N\ \text C^{-1}$

Hence, the electric field intensity at P is $144\left(4\hat{\text j} + 3\hat{\text k}\right)$ N C^-1.

Two point-charges q₁ = + 0.2 C and q₂ = + 0.4 C are 0.1 m apart. Find the electric field at (i) mid-point between the charges, (ii) a point on the line joining q₁ and q₂ such that it is 0.05 m away from q₂ and 0.15 m away from q₁.

Answer

Given,

• q₁ = + 0.2 C, q₂ = + 0.4 C
• Separation = 0.1 m

(i) At the mid-point : The mid-point is 0.05 m from each charge. Both the charges are positive, so their fields there are oppositely directed.

$\text E_1 = \dfrac{\left(9 \times 10^9\right)(0.2)}{(0.05)^2} \\[1em] \text E_1 = \dfrac{1.8 \times 10^9}{2.5 \times 10^{-3}} \\[1em] \text E_1 = 7.20 \times 10^{11}\ \text N\ \text C^{-1}$

directed away from q₁, and

$\text E_2 = \dfrac{\left(9 \times 10^9\right)(0.4)}{(0.05)^2} \\[1em] \text E_2 = \dfrac{3.6 \times 10^9}{2.5 \times 10^{-3}} \\[1em] \text E_2 = 1.44 \times 10^{12}\ \text N\ \text C^{-1}$

directed away from q₂, that is, towards q₁.

$\text E = \text E_2 - \text E_1 \\[1em] \text E = 1.44 \times 10^{12} - 0.72 \times 10^{12} \\[1em] \text E = 7.20 \times 10^{11}\ \text N\ \text C^{-1}$

Hence, the field at the mid-point is 7.20 × 10¹¹ N C^-1, directed towards q₁.

(ii) At a point 0.15 m from q₁ and 0.05 m from q₂ : This point lies on the line joining the charges, beyond q₂. Here both fields are directed away from the charges, that is, in the same direction.

$\text E_1 = \dfrac{\left(9 \times 10^9\right)(0.2)}{(0.15)^2} \\[1em] \text E_1 = \dfrac{1.8 \times 10^9}{2.25 \times 10^{-2}} \\[1em] \text E_1 = 8.0 \times 10^{10}\ \text N\ \text C^{-1} \\[1em] \text E_2 = \dfrac{\left(9 \times 10^9\right)(0.4)}{(0.05)^2} \\[1em] \text E_2 = 1.44 \times 10^{12}\ \text N\ \text C^{-1} \\[1em] \text E = \text E_1 + \text E_2 \\[1em] \text E = 0.08 \times 10^{12} + 1.44 \times 10^{12} \\[1em] \text E = 1.52 \times 10^{12}\ \text N\ \text C^{-1}$

Hence, the field at this point is 1.52 × 10¹² N C^-1, directed away from q₁ and q₂ both.

Electric charges ± 1000 μC are placed at points A and B, respectively at a distance of 2 m from each other. Calculate the electric fields at (i) mid-point of the line AB and (ii) at a point at equal distances of 4 m from each charge.

Answer

Given,

• q_A = + 1000 μC = 1 × 10^-3 C at A, q_B = − 1000 μC = − 1 × 10^-3 C at B
• AB = 2 m

(i) At the mid-point of AB : The mid-point is 1 m from each charge. The field due to the positive charge is directed away from A and that due to the negative charge is directed towards B, so both are along AB and they add up.

$\text E_\text A = \text E_\text B = \dfrac{\left(9 \times 10^9\right)\left(1 \times 10^{-3}\right)}{(1)^2} \\[1em] \text E_\text A = \text E_\text B = 9 \times 10^6\ \text N\ \text C^{-1} \\[1em] \text E = \text E_\text A + \text E_\text B \\[1em] \text E = 1.8 \times 10^7\ \text N\ \text C^{-1}$

Hence, the field at the mid-point is 1.8 × 10⁷ N C^-1, directed along AB.

(ii) At a point equidistant, 4 m, from each charge : Such a point lies on the right bisector of AB, that is, in the broad-side-on position of the dipole. Here r = 4 m and half the dipole length is l = 1 m.

$\text E_\text A = \text E_\text B = \dfrac{\left(9 \times 10^9\right)\left(1 \times 10^{-3}\right)}{(4)^2} \\[1em] \text E_\text A = \text E_\text B = \dfrac{9 \times 10^6}{16} \\[1em] \text E_\text A = \text E_\text B = 5.625 \times 10^5\ \text N\ \text C^{-1}$

The components perpendicular to AB cancel, and the components parallel to AB add up,

$\text E = 2\text E_\text A\cos\theta, \qquad \text{where } \cos\theta = \dfrac{\text l}{\text r} \\[1em] \cos\theta = \dfrac{1}{4} \\[1em] \text E = 2 \times 5.625 \times 10^5 \times \dfrac{1}{4} \\[1em] \text E = 2.8 \times 10^5\ \text N\ \text C^{-1}$

Hence, the field at this point is 2.8 × 10⁵ N C^-1, directed parallel to AB.

A body having an excess of 10⁶ electrons is placed in an electric field of 1000 NC^-1 towards East. Find out the magnitude and direction of the force acting on the body.

Answer

Given,

• Excess of electrons, n = 10⁶
• Charge on one electron, e = 1.6 × 10^-19 C
• Electric field, E = 1000 N C^-1, directed towards East

Charge on the body : An excess of electrons means the body carries a negative charge,

$\text q = -\text{ne} \\[1em] \text q = -10^6 \times 1.6 \times 10^{-19} \\[1em] \text q = -1.6 \times 10^{-13}\ \text C$

Force on the body : Taking the direction of the field, towards East, as positive, E = + 1000 N C^-1,

$\text F = \text{qE} \\[1em] \text F = \left(-1.6 \times 10^{-13}\right)(+1000) \\[1em] \text F = -1.6 \times 10^{-10}\ \text N$

The negative sign shows that the force on the body is directed opposite to the electric field.

Hence, the force is 1.6 × 10^-10 N, directed towards West.

A stationary oil drop between two parallel plates has a charge of 3.2 × 10^-19 C and a weight of 1.6 × 10^-14 N. Find the electric field acting on the drop.

Answer

Given,

• Charge on the oil drop, q = 3.2 × 10^-19 C
• Weight of the drop, mg = 1.6 × 10^-14 N

The drop is stationary between the plates, so the upward electric force on it must balance its weight,

$\text{qE} = \text{mg} \\[1em] \text E = \dfrac{\text{mg}}{\text q} \\[1em] \text E = \dfrac{1.6 \times 10^{-14}}{3.2 \times 10^{-19}} \\[1em] \text E = 5.0 \times 10^{4}\ \text N\ \text C^{-1}$

Hence, the electric field acting on the drop is 5.0 × 10⁴ N C^-1.

A system has two charges q_A = + 0.25 μC and q_B = − 0.25 μC at points A (0, 0, −15 cm) and B (0, 0, +15 cm) respectively. Find the total charge and electric dipole moment of the system.

Answer

Given,

• q_A = + 0.25 μC = + 0.25 × 10^-6 C at A (0, 0, −15 cm)
• q_B = − 0.25 μC = − 0.25 × 10^-6 C at B (0, 0, +15 cm)

Total charge :

$\text q_\text A + \text q_\text B = \left(+0.25\ \mu\text C\right) + \left(-0.25\ \mu\text C\right) \\[1em] \text q_\text A + \text q_\text B = 0$

Hence, the total charge of the system is zero.

Electric dipole moment : The two charges are equal and opposite, so they form an electric dipole. The separation between them is

$2\text l = 15\ \text{cm} + 15\ \text{cm} \\[1em] 2\text l = 30\ \text{cm} \\[1em] 2\text l = 0.30\ \text m \\[1em] \text p = \text q \times 2\text l \\[1em] \text p = \left(0.25 \times 10^{-6}\right)(0.30) \\[1em] \text p = 7.5 \times 10^{-8}\ \text{C-m}$

The dipole moment is directed from the negative charge towards the positive charge, that is, from B (0, 0, + 15 cm) to A (0, 0, − 15 cm).

Hence, the total charge is zero and the electric dipole moment is 7.5 × 10^-8 C-m, directed along the − Z axis.

Two charges of +2.4 μC and −2.4 μC are at a distance 2.5 × 10^-3 m apart from each other. Determine the magnitude of the electric field in the broad-side-on position at a distance of 0.30 m from this dipole. If the dipole be rotated through 90°, then what will be the intensity of the field?

Answer

Given,

• Charges of the dipole, q = 2.4 μC = 2.4 × 10^-6 C
• Separation, 2l = 2.5 × 10^-3 m
• Distance of the point, r = 0.30 m

Electric dipole moment :

$\text p = \text q \times 2\text l \\[1em] \text p = \left(2.4 \times 10^{-6}\right)\left(2.5 \times 10^{-3}\right) \\[1em] \text p = 6.0 \times 10^{-9}\ \text{C-m}$

Field in the broad-side-on position : Since r >> l, the dipole is a short dipole,

$\text E_{\text{equatorial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\text p}{\text r^3} \\[1em] \text E_{\text{equatorial}} = \dfrac{\left(9 \times 10^9\right)\left(6.0 \times 10^{-9}\right)}{(0.30)^3} \\[1em] \text E_{\text{equatorial}} = \dfrac{54}{2.7 \times 10^{-2}} \\[1em] \text E_{\text{equatorial}} = 2.0 \times 10^{3}\ \text N\ \text C^{-1}$

When the dipole is rotated through 90° : The given point, which was on the equatorial line, now lies on the axis of the dipole at the same distance r. Hence

$\text E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\text p}{\text r^3} \\[1em] \text E_{\text{axial}} = 2 \times 2.0 \times 10^{3} \\[1em] \text E_{\text{axial}} = 4.0 \times 10^{3}\ \text N\ \text C^{-1}$

Hence, the intensity is 2.0 × 10³ N C^-1 before rotation and 4.0 × 10³ N C^-1 after rotation.

Two point-charges of +1 μC and −1 μC are kept at a distance of 2 cm. They form an electric dipole. This dipole is situated in a uniform electric field of 1 × 10⁵ N-C^-1, with dipole moment vector $\vec{\text p}$ making an angle of 30° with the field. Calculate (i) the magnitude of electric dipole moment and (ii) the torque acting on the dipole. (iii) When would be the torque maximum? How much?

Answer

Given,

• Charges of the dipole, q = 1 μC = 1 × 10^-6 C
• Separation, 2l = 2 cm = 2 × 10^-2 m
• Electric field, E = 1 × 10⁵ N C^-1
• Angle between $\vec{\text p}$ and $\vec{\text E} \space$, θ = 30°

(i) Magnitude of the electric dipole moment :

$\text p = \text q \times 2\text l \\[1em] \text p = \left(1 \times 10^{-6}\right)\left(2 \times 10^{-2}\right) \\[1em] \text p = 2 \times 10^{-8}\ \text{C-m}$

(ii) Torque acting on the dipole :

$\tau = \text{pE}\sin\theta \\[1em] \tau = \left(2 \times 10^{-8}\right)\left(1 \times 10^{5}\right)\sin 30° \\[1em] \tau = 2 \times 10^{-3} \times \dfrac{1}{2} \\[1em] \tau = 1 \times 10^{-3}\ \text{N-m}$

(iii) Maximum torque :

The torque τ = pE sin θ is maximum when sin θ = 1, that is, when θ = 90°, so that the dipole is normal to the field.

$\tau_{\text{max}} = \text{pE} \\[1em] \tau_{\text{max}} = \left(2 \times 10^{-8}\right)\left(1 \times 10^{5}\right) \\[1em] \tau_{\text{max}} = 2 \times 10^{-3}\ \text{N-m}$

Hence, p = 2 × 10^-8 C-m, τ = 1 × 10^-3 N-m, and the torque is maximum, 2 × 10^-3 N-m, when the dipole is normal to the field (θ = 90°).

An electric dipole, when held at 30° with respect to a uniform electric field of 10⁴ N/C, experiences a torque of 9 × 10^-26 N-m. Calculate the moment of the dipole.

Answer

Given,

• Angle between the dipole and the field, θ = 30°
• Electric field, E = 10⁴ N C^-1
• Torque, τ = 9 × 10^-26 N-m

The torque acting on an electric dipole placed in a uniform electric field is

$\tau = \text{pE}\sin\theta \\[1em] \text p = \dfrac{\tau}{\text E\sin\theta}$

Substituting the values,

$\text p = \dfrac{9 \times 10^{-26}}{\left(10^{4}\right)\left(\sin 30°\right)} \\[1em] \text p = \dfrac{9 \times 10^{-26}}{10^{4} \times \dfrac{1}{2}} \\[1em] \text p = \dfrac{9 \times 10^{-26}}{5 \times 10^{3}} \\[1em] \text p = 1.8 \times 10^{-29}\ \text{C-m}$

Hence, the moment of the dipole is 1.8 × 10^-29 C-m.

Two small identical electrical dipoles AB and CD, each of dipole moment 'p' are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field $\left(\vec{\text E}\right)$ directed along +X direction, what will be the magnitude and direction of the torque acting on this?

Answer

Given,

• Two identical dipoles AB and CD, each of dipole moment p
• Angle between them, 2α = 120°
• Uniform electric field $\vec{\text E} \space$directed along + X

Resultant dipole moment : The two dipole moments are equal in magnitude and are inclined at 120° to each other. Their resultant is

$\text p_{\text{res}} = \sqrt{\text p_{\text{AB}}^2 + \text p_{\text{CD}}^2 + 2\text p_{\text{AB}} \text p_{\text{CD}}\cos 120°} \\[1em] \text p_{\text{res}} = \sqrt{\text p^2 + \text p^2 + 2\text p\text p\cos 120°} \\[1em] \text p_{\text{res}} = \sqrt{2\text p^2 + 2\text p^2\left(-\dfrac{1}{2}\right)} \\[1em] \text p_{\text{res}} = \sqrt{\text p^2} \\[1em] \text p_{\text{res}} = \text p$

Since the two dipole moments are equal, the resultant bisects the angle between them. Using the stated angle of 120°, $\vec{\text p}_{\text{AB}}$ (from B to A) lies along + Y and $\vec{\text p}_{\text{CD}}$ (from D to C) lies at 30° below + X, as in the vector diagram above, which is drawn to this stated angle. Hence the bisector makes an angle of 30° with the + X axis, measured in the anticlockwise direction.

Torque acting on the combination : The angle between $\vec{\text p}_{\text{res}}$ and $\vec{\text E} \space$is θ = 30°,

$\tau = \text p_{\text{res}}\text E\sin\theta \\[1em] \tau = \text{pE}\sin 30° \\[1em] \tau = \dfrac{1}{2}\text{pE}$

The torque $\vec{\tau} = \vec{\text p}_{\text{res}} \times \vec{\text E}$is perpendicular to the plane of the figure. Since $\vec{\text p}_{\text{res}}$ lies in the XY plane above the X-axis and $\vec{\text E} \space$is along + X, the torque is directed along the negative Z-direction.

Hence, the resultant dipole moment is p, at an angle of 30° with the + X axis in the anticlockwise direction, and the torque is $\dfrac{1}{2}\text{pE}$ along the negative Z-direction.

Note: The textbook's sketch is schematic — its line CD is drawn steeper than the marked 120° — so the working uses the stated angle of 120°.