The electrostatic force between two charges of 10 μ C and 20 μ C at a distance of 2m is to be halved. This can be effected by

Question

The electrostatic force between two charges of 10 μ C and 20 μ C at a distance of 2m is to be halved. This can be effected by (A) reducing both charges by 50% (B) reducing both charges and their separation, each by 50% (C) by doubling one of the charges and doubling the separation between them (D) by doubling both charges as well their separation.

Tardigrade Admin · Accepted Answer

Here, F = k (q1q2/d2) and F' = k ((2q1)q2/(2d)2) = (1/2)(k q1 q2/d2) i.e. F' = (F/2)

Q. The electrostatic force between two charges of $10 μ C$ and $20 μ C$ at a distance of $2 m$ is to be halved. This can be effected by

reducing both charges by 50%

reducing both charges and their separation, each by 50%

by doubling one of the charges and doubling the separation between them

by doubling both charges as well their separation.

Here, F = $k \frac{q _{1} q _{2}}{d ^{2}}$ and F' = $k \frac{( 2 q _{1} ) q _{2}}{( 2 d ) ^{2}} = \frac{1}{2} \frac{k q _{1} q _{2}}{d ^{2}} i . e .$ F' = $\frac{F}{2}$

Q. The electrostatic force between two charges of 10μC and 20μC at a distance of 2m is to be halved. This can be effected by