Question

A charge q1 = 5.00nC is placed at the origin of an xy-coordinate system, and a charge q2 = -4.50nC is placed on the positive x axis at x= 4.00cm .

A) If a third charge q3 = 5.00nC is now placed at the point x= 4.00cm , y= 3.00cm find the x and ycomponents of the total force exerted on this charge by the other two charges.

B) Find the magnitude and direction of this force.

Answer 1

Electrostatic force is given by:

F = k*Q1*Q2/R^2

Force between both charge will be attractive if both charge has opposite sign & force between both charge will be repulsive if both charge has same sign.

Given location of each charge is:

Charge A (5.00 nC) is at origin (0, 0)

Charge B (-4.50 nC) is at (4.00, 0) cm

Charge C (5.00 nC) is at (4.00, 3.00) cm

Now we need to find net force on Charge C due to A and B

Step 1. Since charge A and C have Same sign so force between them will be repulsive away from charge C above +ve x-axis,

Fac = k*qa*qc/Rac^2

Rac = distance between Charge A and C = sqrt (4.00^2 + 3.00^2) = 5.00 cm = 0.05 m

Fac = 9*10^9*5.00*10^-9*5.00*10^-9/0.05^2 = 9*10^-5 N

Direction of Force Fac will be

Direction = arctan (3.00/4.00) = 36.87 deg above +ve x-axis (In 1st quadrant)

Now x-component of Force Fac will be:

Fac_x = Fac*cos A = 9*10^-5*cos 36.87 deg

Fac_x = 7.2*10^-5 N

Now y-component of Force Fac will be:

Fac_y = Fac*sin A = 9*10^-5*sin 36.87 deg

Fac_y = 5.4*10^-5 N

Step 2. Since charge B and C have opposite sign so force between them will be attractive towards charge C in -ve y-axis.

Fbc = k*qb*qc/Rbc^2

Rbc = distance between Charge B and C = sqrt (0^2 + 3.00^2) = 3.00 cm = 0.03 m

Fac = 9*10^9*4.50*10^-9*5.00*10^-9/0.03^2 = 22.5*10^-5 N

Since this force is only in -ve y-axis, So

Now x-component of Force Fbc will be:

Fbc_x = 0 N

y-component of Force Fac will be:

Fbc_y = Fbc = -22.5*10^-5 N

Step 3.

Sum of two x-component force will be:

F_x = Fac_x + Fbc_x

F_x = 7.2*10^-5 + 0

F_x = 7.2*10^-5 N

Sum of two y-component force will be:

F_y = Fac_y + Fbc_y

F_y = 5.4*10^-5 - 22.5*10^-5

F_y = -17.1*10^-5 N

Net Force on C will be:

F_net = F_x + F_y

F_net = 7.2*10^-5 i - 17.1*10^-5 j

Magnitude of Net force will be:

|F_net| = 10^-5*sqrt (7.2^2 + (-17.1)^2)

|F_net| = 18.55*10^-5 N

Direction of net force will be:

Above coordinates are in 4th quadrant, so direction will be below the +x-axis

Direction = arctan (F_y/F_x) = arctan (17.1/7.2) = 67.2 deg

Direction = 67.2 deg CW from +ve x-axis = 67.2 deg below the +x-axis = -67.2 deg from +ve x-axis

Let me know if you've any query.