Question

If the potential due to a point charge is 5.05 ✕ 102 V at a distance of 15.7 m, what are the sign and magnitude of the charge? (Enter your answer in C.)

Answer 1

Answer:

The magnitude of charge is $8.809\times 10^{-7}\ C$ with a positive charge.

Explanation:

Given that,

The potential due to a point charge, $V=5.05\times 10^2\ V$

Distance, d = 15.7 m

The potential due to a point charge is given by :

$V=\dfrac{kq}{d}$

q is charge

k is electrostatic constant

$q=\dfrac{Vd}{k}\\\\q=\dfrac{5.05\times 10^2\times 15.7}{9\times 10^9}\\\\q=8.809\times 10^{-7}\ C$  

Here, the potential is positive, then the magnitude of charge is $8.809\times 10^{-7}\ C$ with a positive charge.