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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 F?

A

320 V

B

310 V

C

330 V

D

390 V

Text Solution

Verified by Experts

The correct Answer is:
c
AF = BF
`sqrt((AD)^2 + (DF)^2)`
`= sqrt((4)^2 + (3)^2) = 5 m`
Now potential at point F,
`V_F = 1/(4 pi epsilon_0) [q/(AF) + q/(BF) + (q//2)/(DF) + (q//2)/(CF)]`
`= 1/(4 pi epsilon_0) [q/5 + q/5 + (q//2)/(3) + (q//2)/3]`
`1/(4 pi epsilon_0) xx (11 q)/15 = (9 xx 10^9 xx 11 xx 50 xx 10^(-9))/(15)`
`= 390 V`
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