Question

wo point charges 20 x 10-6 C and -4 X 10-6 C are separated by a distance of 50 cm in air.

(i) Find the point on the line joining the charges, where the electric potential is zero.

(ii) Also find the electrostatic potential energy of the system

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Answer 1

Answer:

20/(50-x)= 4/(x)

5/50-x= 1/x

5x= 50- x

6x=50

x= 8.34

distance = 50-8.34= 41.66

thus

potential= 9×10⁶×[20×10⁶/41.66 + (-4)×10⁶/8.34]

=9×10⁶[0.48-0.479]×10⁶

=9× 10-3 V(ans)

Answer 2

Answer:

Hi brindha josika

Explanation:

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