Question

Two charges are placed along the x axis such that charge A is at -3 m, and has a charge of 2.5e-06 C. Charge B is at 2.2 m, and has a charge of -2.5e-06 C.
What is the x component of the force (in Newtons) felt by charge A due to charge B?
What is the x component of the force (in Newtons) felt by charge B due to charge A?
Suppose a third charge were to be placed at position C, -5.6 m on the x axis. What is the x component of the electric field at position C?

Answer 1

Electrostatic force is given by:

F = k*q1*q2/r^2

Where force between charge will be attractive if both charge has opposite signs and force between them will be repulsive if both charge has same signs

Here given that:

q_A = 2.5*10^-6 C at (-3, 0)

q_B = -2.5*10^-6 C at (2.2, 0)

Force on A due to B will be given by:

F_AB = k*q_A*q_B/R12^2

Since both charge have opposite signs, So force between them will attractive, So force on A will be towards B (In +ve x-axis)

R = distance between qA and qB = 2.2 - (-3) = 5.2 m

So,

F_AB = 9*10^9*2.5*10^-6*2.5*10^-6/5.2^2

F_AB = 2.1*10^-3 N (towards +ve x-axis)

Part B.

From newton's 3rd law every action has equal and opposite reaction, So

Force on A due to B = -Force on B due to A

So Force on B due to A = F_BA = -2.1*10^-3 N (towards -ve x-axis)

OR you can simply use F_BA = k*q_A*q_B/R^2 (which will give same result as above but in opposite direction)

Part C.

Electric field is given by

E = kQ/R^2

We know that electric field due to positive charge is away from the positive charge and towards the negative charge. So at point P (x = -5.6 m) electric field due to each charge will be:

Electric field due to q_B, since q_B is negative, So E_B will be toward q_B (in +ve x-axis)

Electric field due to q_A, since q_A is positive, So E_A will be away from q_A (in -ve x-axis)

Net electric field due to given charges will be

E = E_B - E_A

E = kq_B/r1^2 - kq_A/r2^2

r1 = distance between q_B and P = 2.2 - (-5.6) = 7.8 m

r2 = distance between q_A and P = -3 - (-5.6) = 2.6 m

So,

E = 9*10^9*2.5*10^-6/7.8^2 - 9*10^9*2.5*10^-6/2.6^2

E = -2958.6

E = -2.96*10^3 N/C = x-component of electric field (-ve sign means direction of electric field is towards -ve x-axis)

Since charges are on the x-axis, along with Point P, So electric field will due to the charge will only be on x-axis

Let me know if you've any query.