Question

Two point charges of magnitude 5.0 nC and -3.0 nC are separated by

Answer 1

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