What is the electric force between two points charges when q1=-4e, q2 = +3e, and r = 0.05 m?

Option A. 1.1×10¯²⁴ N

Explanation:

The following data were obtained from the question:

Charge 1 (q1) = - 4e

Charge 2 (q2) = + 3e

Distance apart (r) = 0.05 m

Electric field constant (K) = 9×10⁹ N•m²/C²

Electron (e) = 1.6×10¯¹⁹ C.

Electric Force (F) =..?

Next, we shall determine the value of the two charges.

This is illustrated below:

Charge 1 (q1) = - 4e

Charge 1 (- q1) = 4e

Electron (e) = 1.6×10¯¹⁹ C.

Charge 1 (- q1) = 4 × 1.6×10¯¹⁹ C.

Charge 1 (- q1) = 6.4×10¯¹⁹ C.

Charge 2 (q2) = + 3e

Electron (e) = 1.6×10¯¹⁹ C.

Charge 2 (q2) = 3 × 1.6×10¯¹⁹ C.

Charge 2 (q2) = 4.8×10¯¹⁹ C.

Finally, we shall determine the value of the electric force. This can be obtained as shown below:

Charge 1 (- q1) = 6.4×10¯¹⁹ C.

Charge 2 (q2) = 4.8×10¯¹⁹ C.

Electric field constant (K) = 9×10⁹ N•m²/C².

Distance apart (r) = 0.05 m

Electric Force (F) =..?

F = Kq1q2 /r²

F = (9×10⁹× 6.4×10¯¹⁹× 4.8×10¯¹⁹)/(0.05)²

F = 1.1×10¯²⁴ N

Therefore, the electric force between the two point charge is 1.1×10¯²⁴ N