Question

What is the force F⃗ on the 3.0 nC charge in the middle of (Figure 1) due to the four other charges? Give your answer in component form.

Enter the x and y components of the force separated by a comma.

Answer 1

Electrostatic force is given by:

F = kq1q2/r^2

Since all four charge on corner are same, so due to them force will be equal in magnitude, which will be

|q1| = 2 nC, & |q2| = q = 3.0 nC

r = sqrt (a^2/4 + a^2/4) = a/sqrt 2

a = 1.0 cm = 0.01 m

So,

|F| = 9*10^9*2*10^-9*3.0*10^-9/(0.01/sqrt 2)^2

|F| = 1.08*10^-3 N

Now let's talk about direction, assuming Q is at origin, then

Due to charge in first quadrant, since both charge have opposite signs So force on q will be attractive, towards 45 deg above +ve x-axis in 1st quadrant.

Due to charge in 2nd quadrant, since both charge are negative, So force on q will be repulsive, towards 45 deg below +ve x-axis in 4th quadrant.

Due to charge in 3rd quadrant, since both charge are negative, So force on q will be repulsive, towards 45 deg above +ve x-axis in 1st quadrant.

Due to charge in 4th quadrant, since both charges have opposite signs So the force on q will be attractive, towards 45 deg below +ve x-axis in 4th quadrant.

Now F1 and F3 are above +ve x-axis at 45 deg and F2 and F4 are below +ve axis at 45 deg

F1x = F1*cos 45 deg & F1y = F1*sin 45 deg

F2x = F2*cos 45 deg & F2y = -F2*sin 45 deg

F3x = F3*cos 45 deg & F3y = F3*sin 45 deg

F4x = F4*cos 45 deg & F4y = -F4*sin 45 deg

Since all forces are equal in magnitude, F1 = F2 = F3 = F4 = F

Now add all of them

Fnet = 4*F*cos 45 deg i + 0 j

Sum of all force in y-direction will be zero.

So

Fx_net = 4*1.08*10^-3*cos 45 deg = 3.05*10^-3 N

Fx_net, Fy_net = 3.05*10^-3, 0 N

Direction = +ve x-axis = 0 deg Counter clockwise from +ve x-axis

Let me know if you've any query.