An electric field of $20 N C^{- 1}$ exists along the x-axis in space. Calculate the potential difference $V_{B} - V_{A}$ where the points A and B are given by,
(a) $A = (0, 0); B = (4 m, 2 m)$
(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)?

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Solution

(a) $A = (0, 0), B = (4, 2)$

The displacement in the X- direction is $4 m$ .

The potential difference in the space due to electric field is:
$V_{B} - V_{A} = E \times d$
$= 20 \times 4 = 80 V$

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

The displacement between the two points will be $2 m$ in X-direction.

Therefore, the potential difference is given as:
$\Rightarrow V_{B} - V_{A} = E \times d$
$= 20 \times 2 = 40 V$

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

The displacement in the X- direction is $6 m$ .

The potential difference in the space due to electric field is:
$V_{B} - V_{A} = E \times d$
$= 20 \times 6 = 120 V$

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