Answer :
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:
[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}[/tex]
Applying these equations to the given variables yields
[tex]E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}[/tex]
Divide the first line to the second line:
[tex]\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[/tex]
Using this distance in either of the equations give the magnitude of the charge.
[tex]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[/tex]