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Four charges are placed at the vertices ...

Four charges are placed at the vertices of a rectangle as shown in figure. The magnitude of cha nC. Assume E and F are the mid points of AB and DC respectively.

What is the potential at point E?

A

390 V

B

350 V

C

310 V

D

300 V

Text Solution

Verified by Experts

The correct Answer is:
a
`DE = CE = sqrt((AE)^2 + (AD)^2)`
`= sqrt((3)^2 + (4)^2) = 5 m`
Now , potential at point E,
`V_E = 1/(4 pi epsilon_0) [q/(AE) + q/(BE) + (q//2)/(DE) = (q//2)/(CE)]`
`= 1/(4 pi epsilon_0)[q/3 + q/3 + (q//2)/5 + (q//2)/5]`
`= 1/(4 pi epsilon_0)[(2q)/3 + q/5]`
`= 1/(4pi epsilon_0) xx (13 q)/(15)`
`= (9 xx 10^9 xx 13 xx 50 xx 10^(-9))/(15)`
`= 390 V`
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