Question

A charge q1 = + 10.00 nC is placed at the origin of an xy coordinate system, and a charge q2 = - 10.00 nC is placed on the positive x-axis at x = 8.00 cm. If a third charge q3 = +10.00 nC is now placed at the point x = 8.00 cm, y = 6.00 cm, find the magnitude of the total force exerted on this charge by the other two

Answer 1

Electrostatic force is given by:

F = k*q1*q2/r^2

Net force on charge q3 will be

Fnet = F1 + F2

Since q1 and q3 are positive, So force will be repulsive because to these charges on q3. Since q2 is negative and q3 is positive, force will be attractive.

Direction of F1 = Force due to q1 = at deg above +x-axis

Direction of F2 = force due to q2 = towards -y-axis in (south)

Now

F1 = k*q1*q3/r1^2

here, = arctan(6/8)

r1 = distance between q1 and q3 = sqrt(8.0^2 + 6.0^2) = 10.0 cm = 0.1 m

given, q1 = q3 = 10.0 nC = 10.0*10^-9 C

Using given values:

F1 = 9*10^9*10.0*10^-9*10.0*10^-9/0.1^2 = 9.0*10^-5 N

F2 = (9.0*cos i + 9.0*sin j)*10^-5

F2 = 9.0*8/10 i + 9.0*6/10 j)*10^-5

F2 = 7.2*10^-5 i + 5.4*10^-5 j

Now

F2 = k*q3*q2/r2^2

here, r2 = distance between q2 and q3 = 6.00 cm = 0.06 m

given, q2 = -10.0 nC = -10.0*10^-9 C

Using given values:

F2 = 9*10^9*10.0*10^-9*10.0*10^-9/0.06^2

F2 = -(25*10^-5 N) j

Now net force will be

Fnet = F1 + F2

Fnet = (7.2*10^-5 i + 5.4*10^-5 j) - 25*10^-5 j

Fnet = 7.2*10^-5 i - 19.6*10^-5 j

Magnitude of force will be

|Fnet| = sqrt (Fnet_x^2 + Fnet_y^2)

|Fnet| = sqrt(7.2^2 + 19.6^2)*10^-5

|Fnet| = 20.88*10^-5 N

Let me know if you have any doubt.