Question

An electric field of 20 NC⁻¹ exists along the x-axis in space. Calculate the potential difference V_B − V_A where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?

Answer 1

Given:
Electric field intensity, E = 20 N/C
The electric field is along the x-axis. So, while calculating the potential difference between points B and A using the formula V_B − V_A = E.ds, we will use the difference of the x-coordinates of these point as ds.
(a) A = (0, 0) B = (4 m, 2 m).
So, V_B − V_A = E.ds = 20 × (0 − 4 m) = − 80 V

(b) A = (4 m, 2 m), B = (6 m, 5 m)
⇒ V_B − V_A = E.ds = 20 × (4 − 6) = − 40 V.

(c) A = (0, 0), B = (6 m, 5 m)
⇒ V_B − V_A = E.ds = 20 × (0 − 6) = − 120 V.

Potential difference between points (0, 0) and (6 m, 5 m) = Potential difference between points (0, 0) and (4 m, 2 m) + Potential difference between points (4 m, 2 m) and (6 m, 5 m)