Numerical Problems Based on Coulomb's Law for Class 12 Physics

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Home CBSE Class 12 Physics Numerical Problems Problems on Coulomb’s Law

Coulomb’s Law

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Coulomb’s law, named after French physicist Charles-Augustin de Coulomb, describes the electrostatic interaction between two electrically charged particles.

The law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Mathematically, Coulomb’s law can be expressed as:

F = k q₁ q₂ / r²

where F is the magnitude of the electrostatic force between the two charges, q₁ and q₂ are the magnitudes of the charges, r is the distance between them, and k is Coulomb’s constant, which has a value of approximately 9 × 10⁹ N·m²/C².

Some key features of Coulomb’s law include:

The force is proportional to the product of the charges: The greater the magnitude of the charges, the greater the electrostatic force between them.
The force is inversely proportional to the square of the distance: As the distance between the charges increases, the force between them decreases rapidly.
The force is attractive if the charges are opposite in sign: If the two charges are of opposite signs, the electrostatic force between them is attractive, meaning that the charges will be pulled towards each other.
The force is repulsive if the charges are the same in sign: If the two charges are of the same sign, the electrostatic force between them is repulsive, meaning that the charges will push each other away.

Numerical Problems Based on Coulomb’s Law for Class 12 Physics

Here we are providing numerical problems based on topic Coulomb’s Law for Class 12 Physics.

Numerical Problems on Coulomb’s Law

Class 12 Physics · Electrostatics

Coulomb’s Law Practice Set

Apply the formula $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$ to solve the following

Ch 1 · Electric Charges and Fields

Calculate coulomb force between two $\alpha$-particles separated by a distance of $3.2 \times 10^{-15}$ m in air.

Answer & Solution

Answer: 90 N

Explanation: The charge on an $\alpha$-particle is $+2e$. Therefore, $q_1 = q_2 = 2 \times 1.6 \times 10^{-19}\text{ C} = 3.2 \times 10^{-19}\text{ C}$. The distance $r = 3.2 \times 10^{-15}\text{ m}$.

Using Coulomb’s Law:

$$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

$$F = \frac{(9 \times 10^9) \times (3.2 \times 10^{-19}) \times (3.2 \times 10^{-19})}{(3.2 \times 10^{-15})^2}$$

$$F = 9 \times 10^9 \times 10^{-8} = 90\text{ N}$$

The force is 90 N (repulsive).

The distance between the electron and proton in a hydrogen atom is $5.3 \times 10^{-11}$ m. Determine the magnitude of the ratio of electrostatic and gravitational force between them.

Given: $m_e = 9.1 \times 10^{-31}\text{ kg}$, $m_p = 1.67 \times 10^{-27}\text{ kg}$, $e = 1.6 \times 10^{-19}\text{ C}$ and $G = 6.67 \times 10^{-11}\text{ Nm}^2\text{kg}^{-2}$

Answer & Solution

Answer: $\frac{F_e}{F_G} = 2.27 \times 10^{39}$

Explanation: The electrostatic force $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$ and the gravitational force $F_G = G \frac{m_e m_p}{r^2}$.

Taking the ratio, the $r^2$ term cancels out:

$$\frac{F_e}{F_G} = \frac{\frac{1}{4\pi\epsilon_0} e^2}{G m_e m_p}$$

$$\frac{F_e}{F_G} = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11}) \times (9.1 \times 10^{-31}) \times (1.67 \times 10^{-27})}$$

$$\frac{F_e}{F_G} = \frac{23.04 \times 10^{-29}}{101.3 \times 10^{-69}} \approx 2.27 \times 10^{39}$$

How far apart should two electrons be, if the force each exerts on the other is equal to the weight of the electron?
Given that $e = 1.6 \times 10^{-19}\text{ C}$ and $m_e = 9.1 \times 10^{-31}\text{ kg}$.

Answer & Solution

Answer: 5.08 m

Explanation: For the electrostatic force to equal the weight of the electron, $F_e = W$, which means $\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} = m_e g$.

$$r^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m_e g}$$

Assuming $g = 9.8\text{ m/s}^2$:

$$r^2 = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{9.1 \times 10^{-31} \times 9.8} = \frac{23.04 \times 10^{-29}}{89.18 \times 10^{-31}}$$

$$r^2 \approx 25.83\text{ m}^2$$

$$r = \sqrt{25.83} \approx 5.08\text{ m}$$

Two identical metallic spheres, having unequal, opposite charges are placed at a distance 0.90 m apart in air. After bringing them in contact with each other, they are again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. Calculate the final charge on each of them.

Answer & Solution

Answer: $1.5 \times 10^{-6}\text{ C}$

Explanation: When two identical metallic spheres are brought into contact, their net charge is distributed equally between them. Let the final charge on each sphere be $q$. They now experience a force of repulsion $F = 0.025\text{ N}$ at a distance $r = 0.90\text{ m}$.

$$F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$$

$$0.025 = \frac{9 \times 10^9 \times q^2}{(0.90)^2}$$

$$q^2 = \frac{0.025 \times 0.81}{9 \times 10^9} = \frac{0.02025}{9 \times 10^9} = 2.25 \times 10^{-12}\text{ C}^2$$

$$q = \sqrt{2.25 \times 10^{-12}} = 1.5 \times 10^{-6}\text{ C}$$

Calculate the distance between two protons such that the electrical repulsive force between them is equal to the weight of either.

Answer & Solution

Answer: $11.8\text{ cm}$ (or $0.118\text{ m}$)

Explanation: Here, the electrostatic force equals the weight of a proton. So, $\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} = m_p g$. The mass of a proton $m_p \approx 1.67 \times 10^{-27}\text{ kg}$ and let $g = 9.8\text{ m/s}^2$.

$$r^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m_p g}$$

$$r^2 = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{1.67 \times 10^{-27} \times 9.8} = \frac{23.04 \times 10^{-29}}{16.366 \times 10^{-27}}$$

$$r^2 \approx 1.407 \times 10^{-2}\text{ m}^2$$

Taking the square root:

$$r \approx 0.118\text{ m} = 11.8\text{ cm}$$

Two point charges $A$ and $B$, having charges $+Q$ and $-Q$ respectively, are placed at a certain distance apart and the force acting between them is $F$. If $25\%$ of the charge of $A$ is transferred to $B$, then calculate the new force between the charges in terms of $F$.

Answer & Solution

Answer: $\frac{9}{16}F$

Explanation: Initial force $F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{r^2}$.

When $25\%$ (which is $Q/4$) of charge is transferred from A to B:

New charge on A: $q_A = Q – \frac{Q}{4} = \frac{3Q}{4}$

New charge on B: $q_B = -Q + \frac{Q}{4} = -\frac{3Q}{4}$

The new force $F’$ will be:

$$F’ = \frac{1}{4\pi\epsilon_0} \frac{(\frac{3Q}{4})(\frac{3Q}{4})}{r^2} = \frac{9}{16} \left( \frac{1}{4\pi\epsilon_0} \frac{Q^2}{r^2} \right)$$

$$F’ = \frac{9}{16}F$$

The electrostatic force on a small sphere of charge $0.4\text{ }\mu\text{C}$ due to another small sphere of charge $-0.8\text{ }\mu\text{C}$ in air is $0.2\text{ N}$. What is the distance between the two spheres?

Answer & Solution

Answer: $0.12\text{ m}$ (or $12\text{ cm}$)

Explanation: Given $q_1 = 0.4 \times 10^{-6}\text{ C}$, $q_2 = 0.8 \times 10^{-6}\text{ C}$ (magnitude), and $F = 0.2\text{ N}$.

$$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

$$0.2 = \frac{9 \times 10^9 \times (0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{r^2}$$

$$r^2 = \frac{9 \times 10^9 \times 0.32 \times 10^{-12}}{0.2} = \frac{2.88 \times 10^{-3}}{0.2} = 14.4 \times 10^{-3} = 1.44 \times 10^{-2}\text{ m}^2$$

Taking the square root:

$$r = 0.12\text{ m}$$

Two fixed point charges $+4e$ and $+e$ units are separated by a distance ‘$a$’. Where should a third point charge $q$ be placed on the line joining the two charges for it to be in equilibrium?

Answer & Solution

Answer: At a distance of $\frac{2a}{3}$ from the $+4e$ charge.

Explanation: Let the third charge $q$ be placed at a distance $x$ from the $+4e$ charge. Then its distance from $+e$ is $(a-x)$. For equilibrium, the net force on $q$ must be zero.

$$F_{4e} = F_{e} \implies \frac{1}{4\pi\epsilon_0} \frac{(4e)(q)}{x^2} = \frac{1}{4\pi\epsilon_0} \frac{(e)(q)}{(a-x)^2}$$

$$\frac{4}{x^2} = \frac{1}{(a-x)^2}$$

Taking the square root of both sides:

$$\frac{2}{x} = \frac{1}{a-x}$$

$$2a – 2x = x \implies 3x = 2a \implies x = \frac{2a}{3}$$

The electrostatic force of repulsion between two positively charged ions carrying equal charges is $3.7 \times 10^{-9}\text{ N}$, when they are separated by a distance of $5\text{ \AA}$. How many electrons are missing from each ion?

Answer & Solution

Answer: $2$ electrons

Explanation: Let the charge on each ion be $q$. Given $r = 5\text{ \AA} = 5 \times 10^{-10}\text{ m}$ and $F = 3.7 \times 10^{-9}\text{ N}$.

$$F = 9 \times 10^9 \frac{q^2}{r^2} \implies 3.7 \times 10^{-9} = 9 \times 10^9 \frac{q^2}{(5 \times 10^{-10})^2}$$

$$q^2 = \frac{3.7 \times 10^{-9} \times 25 \times 10^{-20}}{9 \times 10^9} \approx 10.27 \times 10^{-38}\text{ C}^2$$

Taking the square root gives $q \approx 3.2 \times 10^{-19}\text{ C}$.

Since $q = ne$, the number of missing electrons $n$ is:

$$n = \frac{q}{e} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2$$

What is the force between two small charged spheres having charges of $2 \times 10^{-7}\text{ C}$ and $3 \times 10^{-7}\text{ C}$ placed $30\text{ cm}$ apart in air?

Answer & Solution

Answer: $6 \times 10^{-3}\text{ N}$ (repulsive)

Explanation: Given $q_1 = 2 \times 10^{-7}\text{ C}$, $q_2 = 3 \times 10^{-7}\text{ C}$, and $r = 30\text{ cm} = 0.3\text{ m}$.

$$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

$$F = \frac{(9 \times 10^9) \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}$$

$$F = \frac{54 \times 10^{-5}}{0.09} = 600 \times 10^{-5} = 6 \times 10^{-3}\text{ N}$$

Since both charges are positive, the force is repulsive.

Two identical metallic spheres, having charges $+4q$ and $-2q$ are placed at a distance $r$ apart in air. The force of attraction between them is $F$. They are touched together and then placed at the same distance $r$. What is the new force between them?

Answer & Solution

Answer: $\frac{F}{8}$ (repulsive)

Explanation: Initial force $F = k \frac{(4q)(2q)}{r^2} = k \frac{8q^2}{r^2}$ (attractive).

When touched, the net charge $(+4q – 2q = +2q)$ is distributed equally. So, the new charge on each sphere is $+q$.

New force $F’ = k \frac{(q)(q)}{r^2} = k \frac{q^2}{r^2}$ (repulsive).

Comparing the two, we get $F’ = \frac{F}{8}$.

A charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. Find the ratio $Q/q$ so that the force of repulsion between them is maximum.

Answer & Solution

Answer: $2:1$

Explanation: The force between the two parts is $F = k \frac{q(Q-q)}{R^2}$.

For $F$ to be maximum with respect to $q$, the derivative $\frac{dF}{dq}$ must be zero.

$$\frac{d}{dq} [q(Q-q)] = 0 \implies \frac{d}{dq} (qQ – q^2) = 0$$

$$Q – 2q = 0 \implies Q = 2q$$

Therefore, the ratio $\frac{Q}{q} = \frac{2}{1}$.

Two point charges placed at a certain distance $r$ in air exert a force of $40\text{ N}$ on each other. If they are kept at the same distance in water (dielectric constant $K = 80$), calculate the new electrostatic force between them.

Answer & Solution

Answer: $0.5\text{ N}$

Explanation: The force between two charges in a medium with dielectric constant $K$ is given by $F_m = \frac{F_{air}}{K}$.

Given $F_{air} = 40\text{ N}$ and $K = 80$.

$$F_m = \frac{40}{80} = 0.5\text{ N}$$

Calculate the electrostatic force between a proton and an alpha particle separated by a distance of $10^{-13}\text{ m}$ in free space.

Answer & Solution

Answer: $4.6 \times 10^{-2}\text{ N}$

Explanation: Charge on a proton $q_1 = +e$ and charge on an alpha particle $q_2 = +2e$. Distance $r = 10^{-13}\text{ m}$.

$$F = \frac{1}{4\pi\epsilon_0} \frac{(e)(2e)}{r^2} = 9 \times 10^9 \frac{2 \times (1.6 \times 10^{-19})^2}{(10^{-13})^2}$$

$$F = 9 \times 10^9 \frac{2 \times 2.56 \times 10^{-38}}{10^{-26}} = 9 \times 10^9 \times 5.12 \times 10^{-12}$$

$$F = 46.08 \times 10^{-3} = 4.608 \times 10^{-2}\text{ N}$$

If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them?

Answer & Solution

Answer: The force remains unchanged.

Explanation: Initial force $F = k \frac{q_1 q_2}{r^2}$.

If the charges are doubled, new charges are $q_1′ = 2q_1$ and $q_2′ = 2q_2$. If the distance is doubled, new distance is $r’ = 2r$.

New force $F’$ will be:

$$F’ = k \frac{(2q_1)(2q_2)}{(2r)^2} = k \frac{4 q_1 q_2}{4 r^2}$$

The factor of $4$ cancels out in the numerator and denominator, leaving $F’ = k \frac{q_1 q_2}{r^2} = F$.

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