Class 12 Physics Electrostatic Potential and Capacitance: NCERT Solutions for 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
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