Answer:
V = 411.43 V
Explanation:
The two forces as a result of each of the 2 charges are;
F1 = kq1•q/r
F2 = kq2.q/r
Where r = r/2 since we are dealing with potential difference at a point midway between the charges.
q1 = 5 nC = 5 × 10^(-9) C
q2 = 3 nC = 3 × 10^(-9) C
k = 9 × 10^(9) N.m²/C²
r = 35 cm = 0.35m
r/2 = 0.35/2
Thus;
F1 = (9 × 10^(9) × 5 × 10^(-9) × q)/(0.35/2)²
F1 = 1469.39q
F2 = (9 × 10^(9) × 3 × 10^(-9) × q)/(0.35/2)²
F2 = 881.63q
Net force acting midway is;
F_net = F1 + F2
F_net = 1469.39q + 881.63q
F_net = 2351.02q
Now, we know that formula for electric potential is;
V = kq/r
Thus ;
V = Fr/q derived from the earlier equation for force we used.
Where F is F_net.
V = 2351.02q × r/q
V = 2351.02r
Recall that we are dealing with midpoint and r = r/2
Thus;
V = 2351.02 × 0.35/2
V = 411.43 V
