Asked by Anon

Two point charges with q1 = +3 .2×^10−7 C and q2 = −4.8×10^−7 C are initially separated by a distance of r =2 .4×10^−4 m.

(a) How much energy is required to double their separation?
(b) What is the electric potential at the point midway between the two charges when their separation is 2r?

Answered by Arora
a)

Initial potential energy of the system =
k(q1)(q2)/r

Final potential energy of the system =
k(q1)(q2)/2r

Work done = Energy Required = Loss in potential Energy of the system
k(q1)(q2)/r - k(q1)(q2)/2r
= k(q1)(q2)/2r
= (9*10^9)*(3.2*10^-7)*(4.8*10^-7)/4.8*10^-4
= 2.88J
Answered by Arora
b) The potential is a scalar, and can simply be added up.

Total Potential = k(q1)/(r1) + k(q2)/(r2)

At the point midway, both distances are half of 2r, that is, r1 = r2 = r

=> Potential = k((q1)/(r) + (q2)/r))
= (k/r)((3.2 - 4.8)*10^-7)
= -(9*10^9)(1.6*10^-7)/(2.4*10^-4)
= -6*10^6 V
Answered by bobpursley
https://www.youtube.com/watch?v=VmpihFsziFw
Answered by Seb
in a) Why are you dividing by 2r when that change in seperation is r?
There are no AI answers yet. The ability to request AI answers is coming soon!