Answer:
9.6kV
Explanation:
Electric potential is given as
V = k*Q/R
Then the Net electric potential at point A due to given two charges will be:
V_net = V1 + V2
V_net = k*q1/r1 + k*q2/r2
q1 = +4.0*10^-6 C & q2 = -2.0*10^-6 C
r1 = distance between q1 and point A = sqrt (3.0^2 + 4.0^2) = 5.0 m
r2 = distance between q2 and point A = 4.0 - (-1.0) = 5.0 m
Thus,
V_net = k*(q1/r1 + q2/r2)
V_net = 9*10^9*(4.0*10^-6/5.0 - 2.0*10^-6/5.0) = 3600 V
V_net = 3.6 k
To calculate electric potential at origin, then
V0_net = k*q1/R1 + k*q2/R2
R1 = 3.0 m & R2 = 1.0 m, then
V0_net = 9*10^9*(4.0*10^-6/3.0 - 2.0*10^-6/1.0) = -6000 V
generally it is assumed that electric potential at origin is zero, but in this case it's mentioned that we need to recalibrate potential at origin to zero
Now since ask that we need electric potential at point A with relative to zero at the origin, So if we re-calibrate electric potential to zero at the origin, then we need to add 6000 V everywhere, therefore
Net electric potential at Point A = V_net - V0_net = 3600 - (-6000) = 9600 V = 9.6 kV
