Question

A +8.60 μC point charge is sitting at the origin. What is the radial distance between the 500 V equipotential surface and the 1000 Vsurface?

What is the distance between the 1000 Vsurface and the 1500 V surface?

The decrease in spacing of the equipotentials as you move ____________ the charge is due to the fact that the potential is proportional to the ___________ distance, not the distance.

Fill in the blanks with either squared, farther away from, inverse, closer to

Answer 1

Let's consider a point change q = 8 - 6 mu C at the origin, surrounded by their equipotential surfaces 1000v, and 500 v at a distance r_1, and r_2 respectively. Here we have to find out (r_2-r_1) and (r_1-r_2) r_2-r_1=? r_1-r_3=? The potential at a distance r due to point charge q is given by v(r) = 1/ 4 pi epsilon middot q/r Therefore 1000v = 1/4 pi epsilon middot 8.6 mu C/r_1 => r_1 = 1/4 pi epsilon middot 8.6 mu C/1000 v rightarrow 1 similarly r_2 = 1/4 pi epsilon middot 8.6 mu C/500 v rightarrow 2 from 1 and 2 r_2-r_1 = 1/4 pi r_0 8.6 mu C (1/500 v - 1/1000 v) = 9 times 10^9 NM^2/C^2 times 8.6 mu C times (1/1000 v) = 9 times 10^9 NM^2/C^2 times 8.6 times 10^-6 C times 1/1000 NM/C = 77.4 m \r\n Again r_1-r_3 = 1/4 pi epsilon 8.6 mu C (1/1000 v-1/1500v) = 1/4 pi epsilon times 8.6 mu C times 1/3000 v = 9 times 10^9 NM^2/C^2 times 8.6 times 10^- 6