Question

If the potential due to a point charge is 5.75 ✕ 102 V at a distance of 15.6 m, what are the sign and magnitude of the charge? (Enter your answer in C.)

Answer 1

Step-by-step explanation:

The given data is as follows.

Potential (V) =
`5.75 * 10^(2) V`

Distance (r) = 15.6 m

Let us assume that q is the magnitude of charge. Formula to calculate charge will be as follows.

V =
`(1)/(4 \pi \epsilon_(o)) * (q)/(r)`

`5.75 * 10^(2) V = 9 * 10^(9) * (q)/(15.6 m)`

q =
`(89.7 * 10^(2))/(9 * 10^(9))`

=
`9.96 * 10^(-7)` C

As the potential due to charge is positive. Therefore, q is positive.

Thus, we can conclude that the sign of charge is positive and magnitude of the charge is
`9.96 * 10^(-7)` C.