First slide

Electrostatic potential

Question

P is a point at a distance r₁ from a point charge Q and potential at P is found to be 5 volt.. Now another point charge 4Q is placed at a distance r₂ from P such that intensity of electric field at P becomes zero. Then final potential at P will be

Moderate

A

10 volt
B

12.5 volt
C

7.5 volt
D

15 volt

Solution

Initially $\frac{K . Q}{r_{1}} = V .$
Now, $\frac{KQ}{r_{_{1}}^{2}} = \frac{K . 4 Q}{r_{2}^{2}} \Rightarrow \frac{r_{2}}{r_{1}} = \sqrt{\frac{4 Q}{Q}} = 2 .$
$∴ V^{|} = \frac{KQ}{r_{1}} + \frac{K . 4 Q}{r_{2}} = \frac{KQ}{r_{1}} (1 + 4 . \frac{r_{1}}{r_{2}}) = V (1 + 4 . \frac{1}{2}) = 3 V \Rightarrow V^{|} = 3 \times 5 volt = 15 volt .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App