Question

(Figure 1) shows five electric charges Four charges with the magnitude of the charge 2.0 nC form a square with the size a = 4.0 cm. Positive charge with the magnitude of q = 4.0 nC is placed in the center of the square. 

Part A

What is the magnitude of the force on the 4.0 nC charge in the middle of the figure due to the four other charges? 

Part B 

What is the direction of the force on the 4.0 nC charge in the middle of the figure due to the four other charges?

Answer 1

Electrostatic force is given by:

F = k*q1*q2/r^2

Force will be attractive if both charge have opposite signs and force will be repulsive if both charges has same sign,

In given Situation: q1 = -2.0 nC at bottom left corner and q2 = -2.0 nC at top left corner

q3 = 2.0 nC at bottom right corner & q4 = 2.0 nC at top right corner

Now we need find net force on charge q at the center of square, So

Force due to charge q1 on q will be attractive and at an angle of 45 deg below -ve x-axis (In 3rd quadrant x < 0 & y < 0)

Force due to charge q2 on q will be attractive and at an angle of 45 deg above -ve x-axis (In 2nd quadrant x < 0 & y > 0)

Force due to charge q3 on q will be repulsive and at an angle of 45 deg above -ve x-axis (In 2nd quadrant x < 0 & y > 0)

Force due to charge q4 on q will be repulsive and at an angle of 45 deg below -ve x-axis (In 3rd quadrant x < 0 & y < 0)

Now since given that: |q1| = |q2| = |q3| = |q4| = 2.0 nC = 2.0*10^-9 C, and all charges are at same distance from the center of square, So |F1| = |F2| = |F3| = |F4|, where

|F1| = k*q1*q/r^2

r = distance between q1 and q = a/sqrt 2 = 4.0/sqrt 2 = 2.828 cm = 0.02828 m

|F1| = 9*10^9*2.0*10^-9*4.0*10^-9/0.02828^2 = 9.00*10^-5 N

Now component of each force in x and y direction will be:

F1 = F1x i + F1y j = -9.00*10^-5*cos 45 deg j - 9.00*10^-5*sin 45 deg j = -6.364*10^-5 i - 6.364*10^-5 j

F2 = F2x i + F2y j = -9.00*10^-5*cos 45 deg j + 9.00*10^-5*sin 45 deg j = -6.364*10^-5 i + 6.364*10^-5 j

F3 = F3x i + F3y j = -9.00*10^-5*cos 45 deg j + 9.00*10^-5*sin 45 deg j = -6.364*10^-5 i + 6.364*10^-5 j

F4 = F4x i + F4y j = -9.00*10^-5*cos 45 deg j - 9.00*10^-5*sin 45 deg j = -6.364*10^-5 i - 6.364*10^-5 j

Now Net force in x direction will be

Fx = F1x + F2x + F3x + F4x = 4*F1x

Fx = -4*6.364*10^-5 N

In two significant figures

Fx = -2.5*10^-4 N

Similarly, Now Net force in y direction will be

Fy = F1y + F2y + F3y + F4y = 0 (Since F1y and F4y are negative & F2y and F3y are positive)

Fy = 0 N

Now Net force will be

Fnet = Fx + Fy

Fnet = -2.5*10^-4 i + 0 j

Fnet = (-2.5*10^-4 N) i

Magnitude of net force will be

|Fnet| = sqrt (Fx^2 + Fy^2)

|Fnet| = sqrt ((-2.5*10^-4)^2 + 0^2)

|Fnet| = 2.5*10^-4 N

Part B.

Direction of net force will be:

Direction = 180 deg Counterclockwise from the +ve x-axis, since net force is in -ve x-axis

Let me know if you've any query.