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sergey [27]
3 years ago
9

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)
Physics
1 answer:
Finger [1]3 years ago
8 0

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

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