Question

A -3.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

Q1: Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 1.20 mm.

Q2: Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = -0.200 m

Answer 1

electrostatic force is given by

F = kQ1*Q2/d^2

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs.

Part A

Net force on electron (q3 = -e) placed at x = 1.20 m

Force F1, on q3 due to q1 will be repulsive and towards the +ve x-axis

Force F2, on q3 due to q2 will be repulsive and towards the +ve x-axis

So net force on q3 will be

Fnet = F1 + F2 = F1 + F2

Fnet = k*q1*q3/d1^2 + k*q2*q3/d2^2

d1 = distance between q1 and q3 = 1.200 m

d2 = 1.200 - 0.800 = 0.400 m

q1 = -3.00 nC = -3.00*10^-9 C

q2 = -5.50 nC = -5.50*10^-9 C

q3 = -e = -1.6*10^-19 C

So Using these values:

Fnet = k*q3*[q1/d1^2 + q2/d2^2]

Fnet = 9*10^9*1.6*10^-19*[3.00*10^-9/1.200^2 + 5.50*10^-9/0.400^2]

Fnet = 0.525*10^-16 N

(+ve sign means net force will be towards right)

Part B.

Net force on electron (q3 = -e) placed at x = -0.200 m

Force F1, on q3 due to q1 will be repulsive and towards the -ve x-axis

Force F2, on q3 due to q2 will be repulsive and towards the -ve x-axis

So net force on q3 will be

Fnet = -F1 - F2 = -(F1 + F2)

Fnet = -(k*q1*q3/d1^2 + k*q2*q3/d2^2)

d1 = distance between q1 and q3 = 0.200 m

d2 = 0.200 + 0.800 = 1.000 m

q1 = -3.00 nC = -3.00*10^-9 C

q2 = -5.50 nC = -5.50*10^-9 C

q3 = -e = -1.6*10^-19 C

So Using these values:

Fnet = -k*q3*[q1/d1^2 + q2/d2^2]

Fnet = -9*10^9*1.6*10^-19*[3.00*10^-9/0.200^2 + 5.50*10^-9/1.000^2]

Fnet = -1.16*10^-16 N

(-ve sign means net force will be towards left)

Let me know if you've any query