Question

find the minimum force in between two charges are separated by 1 meter apart.​

Answer 1

Answer:

We know that, F=(1/4πε0)(q^1q^2)/r^2. F will be minimum when q1andq2 are minimum i. e. q1=q2=1.6×10−19. Then, Fmin=9×10^9N×(1.6×10^−10)^2. Fmin=2.3×10^−28N.

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Answer 2

Answer:

F=(1/4πε0)(q^1q^2)/r^2.

F will be minimum when q1andq2 are minimum i. e. q1=q2=1.6×10−19.

F = kq1×q2/r to the power 2

smallest value of q is 1.6 × 10 to the power -19 C

k = 9× 10 to the power 9 N/C

R = 1m

Fmin=2.3×10 to the power −28N.

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