Numerical Problems Based on Superposition Principle of Electric Forces for Class 12 Physics

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Home CBSE Class 12 Physics Numerical Problems Superposition Principle of Electric Forces

Principle of superposition of electrostatic forces.

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Coulomb’s law gives force between two point charges. The principle of superposition enables us to find the force on a point charge due to a group of point charges. This principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of other charges.

The principle of superposition states that when a number of charges are interacting, the total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force between two charges is not affected by the presence of other charges.

Mathematically, the superposition principle can be expressed as follows:

F_total = F₁ + F₂ + … + F_n

where F_total is the total force on the charged particle, F₁, F₂, …, F_n are the individual forces exerted on the particle by each of the n other charged particles in the system.

Here we are providing numerical problems based on Superposition Principle of Electric Forces for Class 12 Physics.

Numerical Problems on Superposition Principle of Electric Forces

Class 12 Physics · Electrostatics

Superposition Principle Practice Set

Apply vector addition to Coulomb’s Law: $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + … + \vec{F}_n$

Ch 1 · Electric Charges and Fields

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 $1.0 \times 10^{-8}$ C, the second $8 \times 10^{-8}$ C, third $27 \times 10^{-8}$ C, and so on. The tenth particle has a charge $1000 \times 10^{-8}$ C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer & Solution

Answer: $4.95 \times 10^5$ N

Explanation: The charges follow the pattern $q_n = n^3 \times 10^{-8}\text{ C}$. Their positions are $r_n = n \times 10\text{ cm} = n \times 0.1\text{ m}$. The test charge at the origin is $Q = 1\text{ C}$.

According to the superposition principle, the total force is the sum of individual forces. The force from the $n$-th charge is:

$$F_n = \frac{1}{4\pi\epsilon_0} \frac{Q q_n}{r_n^2}$$

$$F_n = \frac{(9 \times 10^9) \times (1) \times (n^3 \times 10^{-8})}{(n \times 0.1)^2}$$

$$F_n = \frac{90 \times n^3}{n^2 \times 0.01} = 9000 n$$

The total force $F$ is the sum from $n=1$ to $10$:

$$F = \sum_{n=1}^{10} 9000 n = 9000 (1 + 2 + 3 + … + 10)$$

$$F = 9000 \times \left(\frac{10 \times 11}{2}\right) = 9000 \times 55 = 495000\text{ N}$$

So, the magnitude of the electric force is $4.95 \times 10^5\text{ N}$.

Charges of +5 μC, +10 μC and -10 μC are placed in air at the corners A, B and C of an equilateral triangle ABC, having each side equal to 5 cm. Determine the resultant force on the charge at A.

Answer & Solution

Answer: 180 N

Explanation: Let $q_A = +5 \times 10^{-6}\text{ C}$, $q_B = +10 \times 10^{-6}\text{ C}$, and $q_C = -10 \times 10^{-6}\text{ C}$. The side length is $a = 0.05\text{ m}$.

Force on A due to B ($F_{AB}$) is repulsive (directed along BA extended):

$$F_{AB} = \frac{(9 \times 10^9)(5 \times 10^{-6})(10 \times 10^{-6})}{(0.05)^2} = \frac{0.45}{0.0025} = 180\text{ N}$$

Force on A due to C ($F_{AC}$) is attractive (directed from A to C):

$$F_{AC} = \frac{(9 \times 10^9)(5 \times 10^{-6})(10 \times 10^{-6})}{(0.05)^2} = 180\text{ N}$$

The angle between $F_{AB}$ and $F_{AC}$ is $120^\circ$ (since the interior angle of the equilateral triangle is $60^\circ$). The resultant force $F_R$ is:

$$F_R = \sqrt{F_{AB}^2 + F_{AC}^2 + 2F_{AB}F_{AC}\cos(120^\circ)}$$

$$F_R = \sqrt{180^2 + 180^2 + 2(180)(180)(-0.5)}$$

$$F_R = \sqrt{180^2} = 180\text{ N}$$

Charges $q_1 = 1.5\text{ mC}$, $q_2 = 0.2\text{ mC}$ and $q_3 = -0.5\text{ mC}$ are placed at the points A, B and C respectively. If $r_1 = 1.2\text{ m}$ (distance AB) and $r_2 = 0.6\text{ m}$ (distance BC), forming a right angle at B, calculate the magnitude of resultant force on $q_2$.

Answer & Solution

Answer: $3.125 \times 10^3\text{ N}$

Explanation: Assuming the charges form a right-angled triangle at B, the forces $F_{12}$ and $F_{32}$ acting on $q_2$ will be perpendicular to each other.

Force on $q_2$ due to $q_1$ ($F_{12}$, repulsive):

$$F_{12} = \frac{k q_1 q_2}{r_1^2} = \frac{(9 \times 10^9)(1.5 \times 10^{-3})(0.2 \times 10^{-3})}{(1.2)^2}$$

$$F_{12} = \frac{2700}{1.44} = 1875\text{ N}$$

Force on $q_2$ due to $q_3$ ($F_{32}$, attractive):

$$F_{32} = \frac{k |q_3| q_2}{r_2^2} = \frac{(9 \times 10^9)(0.5 \times 10^{-3})(0.2 \times 10^{-3})}{(0.6)^2}$$

$$F_{32} = \frac{900}{0.36} = 2500\text{ N}$$

Since the forces are perpendicular, the resultant force $F$ is:

$$F = \sqrt{F_{12}^2 + F_{32}^2} = \sqrt{(1875)^2 + (2500)^2}$$

$$F = \sqrt{3515625 + 6250000} = \sqrt{9765625} = 3125\text{ N} = 3.125 \times 10^3\text{ N}$$

Three point charges +q each are kept at the vertices of an equilateral triangle of side $l$. Determine the magnitude and sign of the charge to be kept at its centroid so that the charges at the vertices remain in equilibrium.

Answer & Solution

Answer: $-q/\sqrt{3}$

Explanation: Let the vertices be A, B, and C, each with charge $+q$. Let the centroid be O with charge $Q$. For any corner charge (e.g., at A) to be in equilibrium, the net force on it must be zero.

The repulsive forces from B and C on A are equal in magnitude:

$$F_B = F_C = \frac{k q^2}{l^2}$$

The angle between them is $60^\circ$. Their resultant $F_R$ is directed outward along the median:

$$F_R = \sqrt{F_B^2 + F_C^2 + 2F_B F_C \cos(60^\circ)} = \sqrt{3F_B^2} = \sqrt{3}\frac{k q^2}{l^2}$$

To balance this, the force from the centroid charge $Q$ must be attractive (meaning $Q$ must be negative) and equal in magnitude to $F_R$. The distance from a vertex to the centroid is $r = l/\sqrt{3}$.

The attractive force from $Q$ is:

$$F_Q = \frac{k |Q| q}{(l/\sqrt{3})^2} = \frac{3k |Q| q}{l^2}$$

Equating $F_Q$ and $F_R$ for equilibrium:

$$\frac{3k |Q| q}{l^2} = \sqrt{3}\frac{k q^2}{l^2}$$

$$3 |Q| = \sqrt{3} q \implies |Q| = \frac{\sqrt{3}}{3} q = \frac{q}{\sqrt{3}}$$

Since the charge must be negative to attract, $Q = -q/\sqrt{3}$.

Superposition Principle Practice Set (Contd.)

Apply vector addition to Coulomb’s Law: $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + … + \vec{F}_n$

Ch 1 · Electric Charges and Fields

Four equal point charges, $+q$ each, are placed at the four corners of a square of side $a$. Find the magnitude of the net electrical force acting on any one of the charges.

Answer & Solution

Answer: $\frac{1}{4\pi\epsilon_0} \frac{q^2}{a^2} \left(\sqrt{2} + \frac{1}{2}\right)$

Explanation: Let the square be ABCD with charges $+q$ at each corner. We will find the force on the charge at corner C.

The force from B ($F_{BC}$) and the force from D ($F_{DC}$) are perpendicular to each other. Their distance is $a$.

$$F_{BC} = F_{DC} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{a^2} = F$$

The resultant of these two perpendicular forces acts along the diagonal AC:

$$F_1 = \sqrt{F^2 + F^2} = F\sqrt{2} = \frac{1}{4\pi\epsilon_0} \frac{q^2\sqrt{2}}{a^2}$$

The force from the charge at corner A ($F_{AC}$) also acts along the diagonal AC. The distance AC is $a\sqrt{2}$.

$$F_{AC} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{(a\sqrt{2})^2} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{2a^2}$$

Since $F_1$ and $F_{AC}$ are in the exact same direction, the total net force is their sum:

$$F_{net} = F_1 + F_{AC} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{a^2}\sqrt{2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{2a^2}$$

$$F_{net} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{a^2} \left(\sqrt{2} + \frac{1}{2}\right)$$

Two point charges $+4e$ and $+e$ are separated by a distance $a$. Where should a third point charge $q$ be placed on the line joining them so that it is in equilibrium?

Answer & Solution

Answer: At a distance of $2a/3$ from the $+4e$ charge.

Explanation: Let the third charge $q$ be placed at a distance $x$ from the $+4e$ charge. Its distance from the $+e$ charge will be $(a – x)$.

For $q$ to be in equilibrium, the force exerted by $+4e$ must be equal and opposite to the force exerted by $+e$.

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

Canceling common terms gives:

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

Taking the square root of both sides:

$$\frac{2}{x} = \frac{1}{a – x}$$

$$2(a – x) = x \implies 2a – 2x = x \implies 3x = 2a$$

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

Three charges $q_1 = 1\text{ }\mu\text{C}$, $q_2 = -2\text{ }\mu\text{C}$, and $q_3 = 3\text{ }\mu\text{C}$ are placed on the x-axis at $x = 0$, $x = 1\text{ m}$, and $x = 2\text{ m}$ respectively. Find the magnitude and direction of the net force on $q_2$.

Answer & Solution

Answer: $36 \times 10^{-3}\text{ N}$ towards the positive x-axis.

Explanation: The charge $q_2$ is negative, while $q_1$ and $q_3$ are positive. Both forces will be attractive.

Force on $q_2$ due to $q_1$ is attractive, directed towards $x=0$ (negative x-direction):

$$F_{12} = \frac{(9 \times 10^9) \times (1 \times 10^{-6}) \times (2 \times 10^{-6})}{1^2} = 18 \times 10^{-3}\text{ N}$$

Force on $q_2$ due to $q_3$ is attractive, directed towards $x=2$ (positive x-direction):

$$F_{32} = \frac{(9 \times 10^9) \times (3 \times 10^{-6}) \times (2 \times 10^{-6})}{1^2} = 54 \times 10^{-3}\text{ N}$$

Since the forces are in opposite directions, we subtract them to find the net force:

$$F_{net} = 54 \times 10^{-3} – 18 \times 10^{-3} = 36 \times 10^{-3}\text{ N}$$

The net force is in the direction of the larger force, which is towards the $+3\text{ }\mu\text{C}$ charge (positive x-axis).

Consider three point charges $q_1, q_2, q_3$ each equal to $q$ placed at the vertices of an equilateral triangle of side $l$. What is the net electrostatic force on a test charge $Q$ placed at the centroid of the triangle?

Answer & Solution

Answer: Zero

Explanation: Let the vertices be A, B, and C, and the centroid be O. The distance from each vertex to the centroid is equal, let’s call it $r$ (where $r = l/\sqrt{3}$).

The forces exerted by the three charges on $Q$ are equal in magnitude because the charges and distances are identical:

$$F_A = F_B = F_C = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r^2}$$

These three force vectors point outward from the centroid to the vertices (if $q$ and $Q$ have opposite signs, they point inward). In either case, the angle between any two of these force vectors is $120^\circ$.

By the principle of vector addition (or Lami’s theorem), the vector sum of three equal forces acting at a point with an angle of $120^\circ$ between each pair is exactly zero.

$$\vec{F}_{net} = \vec{F}_A + \vec{F}_B + \vec{F}_C = 0$$

Two identical point charges $Q$ are kept at a distance $r$ from each other. A third point charge $q$ is placed on the line joining the two charges such that the entire system of three charges is in equilibrium. What is the magnitude, sign, and position of the charge $q$?

Answer & Solution

Answer: Position: exactly in the middle ($r/2$). Magnitude: $Q/4$. Sign: Negative (opposite to $Q$).

Explanation: For $q$ to be in equilibrium between two identical charges, it must be placed exactly at the midpoint. So, distance from each $Q$ is $r/2$.

For the *system* to be in equilibrium, the extreme charges $Q$ must also experience zero net force. Let’s look at one of the $Q$ charges. It experiences a repulsive force from the other $Q$ charge. Therefore, the force from $q$ must be attractive to cancel it out. This means $q$ must be of the opposite sign (negative).

Equating the magnitudes of the forces on one of the $Q$ charges:

$$F_{Q\text{ due to }q} = F_{Q\text{ due to other }Q}$$

$$\frac{1}{4\pi\epsilon_0} \frac{Q|q|}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{r^2}$$

$$\frac{4Q|q|}{r^2} = \frac{Q^2}{r^2}$$

Canceling common terms:

$$4|q| = Q \implies |q| = \frac{Q}{4}$$

Since the sign must be negative, $q = -Q/4$.

Four point charges $q_A = 2\text{ }\mu\text{C}$, $q_B = -5\text{ }\mu\text{C}$, $q_C = 2\text{ }\mu\text{C}$, and $q_D = -5\text{ }\mu\text{C}$ are located at the corners of a square ABCD of side 10 cm. What is the net electrical force on a charge of $1\text{ }\mu\text{C}$ placed at the centre of the square?

Answer & Solution

Answer: Zero

Explanation: The center of the square (O) is equidistant from all four corners. Let this distance be $r$.

The charge at the center is $q_0 = 1\text{ }\mu\text{C}$.

Force due to $q_A$ ($2\text{ }\mu\text{C}$) pushes $q_0$ away along the diagonal AC. Force due to $q_C$ ($2\text{ }\mu\text{C}$) pushes $q_0$ away along the same diagonal AC but in the exact opposite direction. Since both charges are equal in magnitude and sign, $\vec{F}_A$ and $\vec{F}_C$ cancel each other out completely.

Similarly, force due to $q_B$ ($-5\text{ }\mu\text{C}$) pulls $q_0$ towards B along the diagonal BD. Force due to $q_D$ ($-5\text{ }\mu\text{C}$) pulls $q_0$ towards D along the same diagonal but in the opposite direction. $\vec{F}_B$ and $\vec{F}_D$ are equal in magnitude and opposite in direction, so they also cancel each other out.

$$\vec{F}_{net} = (\vec{F}_A + \vec{F}_C) + (\vec{F}_B + \vec{F}_D) = 0 + 0 = 0$$