1.

What is the electric potential at a distance of 9 cm from 3 nC ?

Answer»

270 V
3 V
300 V
30 V

Solution :`V = 9 xx 10^(9) xx (q)/(r) ""(because V = (1)/(4pi epsilon_(0))(q)/(r))`
`V = (9 xx 10^(9) xx 3 xx 10^(-9))/(9 xx 10^(-2)) = 300 V`


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