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

1. d = 0.415 m.

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

Step-by-step explanation:

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

`E = (1)/(4\pi\epsilon_0)(Q)/(r^2)\\V = (1)/(4\pi\epsilon_0)(Q)/(r)`

Applying these equations to the given variables yields

`E = 12 = (1)/(4\pi\epsilon_0)(Q)/(d^2)\\V = 4.98 = (1)/(4\pi\epsilon_0)(Q)/(d)`

Divide the first line to the second line:

`(12)/(4.98) = ( (1)/(4\pi\epsilon_0)(Q)/(d^2))/((1)/(4\pi\epsilon_0)(Q)/(d))\\(12)/(4.98) = (1)/(d)\\d = 0.415~m`

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

`12 = (1)/(4\pi\epsilon_0)(Q)/((0.415)^2)\\12 = (1)/(4\pi (8.8* 10^(-12)))(Q)/((0.415)^2)\\Q = 2.285 * 10^(-10)~C`