Question -
Answer -
Two charges placed at points A and B are represented inthe 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 arethe electric potential and electric field respectively at O.
V1 = Potential due to charge at A + Potential due tocharge at B
Where,
∈0 = Permittivityof free space
E1 = Electric field due to q2 −Electric field due to q1
Therefore, the potential at mid-point is2.4 × 105 V and the electric field at mid-point is 4× 105 Vm−1. The field is directed from the larger charge to the smallercharge.
(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 electricpotential and electric field respectively at Z.
It can be observed from the figure that distance,
V2= Electric potential due to A + Electric Potential due toB
Electric field due to q atZ,
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 electricfield is 6.6 ×105 V m−1.