Question

Two charges of 1C and 4C are placed 0.5m apart. the ratio of the coulomb forces experienced by the two charges is

Answer 1

$q1 = 1c \\ \\ q2 = 4c \\ \\ r = 0.5m \\ \\ force = \frac{q1q2}{ {r}^{2} } \\ \\ = \frac{1 \times 4}{ {0.5}^{2} } \\ \\ = \frac{4}{0.25} \\ \\ = 16$

Answer 2

Answer:

1:1

Explanation:

F1 =k (q1q2/r2)

F2=k (q1q2/r2)

Thus, same force will be acted on them. So, ratio is 1:1.