Question

A charge-50c has electric field and electric potential at a distance 3mt, in air medium?​

Answer 1

Explanation:

as E = kq/r2

$9 \times 10 {} {}^{9} \times 50 \div 3 {}^{2} = 5 \times 10 {}^{10}$

V = kq/r

$9 \times 10 {}^{9} \times 50 \div 3 = 15 \times 10 {}^{10}$

Answer 2

As E =kg/r2
9 x 10^9x50 divided by3=5 x 10^10



V=kg/r9 x 10^9
9 x 10^9 x 50 divided by 3= 15 x 10^10