Question

Two charges +1µc and +4µc are situated at a distance in air. The ratio of the forces acting on them is: A. 1:4 B. 4:1 C. 1:1 D. 1:16​

Answer 1

Answer:

your answer is c.1.1

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Answer 2

Answer:

C - 1:1

Explanation:

q1 = +1µc

q2 = +4µc

We know that,

F = kq1q2/$r^{2}$

Therefore, $F_{12}$ = k x 1µc x 4µc/ $r^{2}$             (1)

and $F_{21}$ = k x 4µc x 1µc/$r^{2}$                         (2)

Dividing (1) by (2),

$\frac{F_{12}}{F_{21}}$ = $\frac{k X 1µc X 4µc/ r^{2}}{k X 4µc X 1µc/r^{2}}$  

$\frac{F_{12}}{F_{21}}$ = 4µc/4µc

$\frac{F_{12}}{F_{21}}$ = 1/1

Therefore, ratio is 1:1.