# Class 12 Physics Electrostatic Potential and Capacitance: NCERT Solutions for Question 14

This page focuses on the detailed Electrostatic Potential and Capacitance question answers for Class 12 Physics Electrostatic Potential and Capacitance, addressing the question: 'Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and   (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
Question 14

## Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:(a) at the mid-point of the line joining the two charges, and(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, q1 = 1.5 μC

Magnitude of charge located at B, q2 = 2.5 μC

Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V1 and E1 are the electric potential and electric field respectively at O

V1 = Potential due to charge at A + Potential due to charge at B

Where, ∈0 = Permittivity of free space

E1 = Electric field due to q2 - Electric field due to q1

Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4 × 105 V m - 1. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

V2 and E2 are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

V2= Electric potential due to A + Electric Potential due to B

Electric field due to q at Z,

Electric field due to q2 at Z,

The resultant field intensity at Z,

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V and electric field is 6.6 ×105 V m -1