Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m, respectively. (Take V = 0 at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

Answer 1

Answer:

A. 30.7cm

B. $1.7*10^{-10}C$

C. The electric field is directed away from the point of charge

Explanation:

A.

 $because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm$

B.

Considering Gauss's law

$EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C$

C. The electric field directed away from the point of charge when the charge is positive.