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Academic session 2026-27, Chapter 2 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Electrostatic Potential and Capacitance. This page offers a step-by-step solution to the specific question from Exercise 1, 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.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 14: Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potentia....
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