Question

An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA 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

Answer: well there is no relation except them being the multiples of 40.

Explanation:

Answer 2

Using $V_B - V_A = Eds$

Explanation:

Given:

Electric field intensity, E = $20\ NC^-^1$

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, $V_B-V_A = E.ds = 20 \times (0 - 4m)$

= - 80 V

(b.) A = (4 m, 2 m), B = (6 m, 5 m)

So, $V_B-V_A = E.ds = 20 \times (4 - 6)$

= - 40 V

(c.) A = (0, 0), B = (6 m, 5 m)

So, $V_B-V_A = E.ds = 20 \times (0 - 6)$

= - 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)