Question

A charge of 25C is moved from infinity to point A and B in an electric field when the work done to do so is 10J and 10.5J respectively. Calculate the potential difference between the points A and B. ​

Answer 1

Answer:

potential difference between two points = Workdone to move unit charge between these two points.

hence potential difference between A and B = (10.5 -10)/25 = 0.02 volt

Answer 2

Solution

Work done by A or $W_{a}$ = 10 J

Work done by B or $W_{b}$ = 10.5 J

Charge "Q" = 25 C

So, Potential at point A

$P_1\: = \: \dfrac{W_a}{Q}$

$P_1\: = \: \dfrac{10}{25}$

$P_1\: = \: \dfrac{2}{5}$

$P_2\: = \: 0.40 \: V$

Now for point B

$P_2\: = \: \dfrac{W_b}{Q}$

$P_2\: = \: \dfrac{10.5}{25}$

$P_2\: = \: \dfrac{105}{250}$

$P_2\: = \: 0.42 \: V$

Potential difference between between A & B

$P \: = \: P_{2}\: - \: P_{1}$

$P \: = \: 0.42\: - \:0.40$

$P\: = \: 0.02\: Volt$