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

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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i really dont know but i will try okay.

so i think u would just add all them up and get your answer..