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?
(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?
Answers
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
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
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?