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
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?
4 answers
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
https://www.youtube.com/watch?v=VmpihFsziFw
in a) Why are you dividing by 2r when that change in seperation is r?
2 answers