Question

At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m, respectively. (Take the potential to be zero at infinity.)1.What is the distance to the point charge? (d= ? m)2.What is the magnitude of the charge? (q= ? c)

Answer 1

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$

Applying these equations to the given variables yields

$E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}$

Divide the first line to the second line:

$\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m$

Using this distance in either of the equations give the magnitude of the charge.

$12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C$