Question

The force acting between two opposite charges of equal magnitude is F. If $25\%$ charges are transferred from one to another, the force acting between them will be____.

• A. $F$
• B. $\frac{4F}{5}$
• C. $\frac{15F}{16}$
• D. $\frac{9F}{16}$

Answer 1

The correct option is C $\frac{15F}{16}$

The force acting between two opposite charge of equal magnitude is

$F=\frac{1}{4\pi {\epsilon }_0}.\frac{q^2}{r^2}$. . . . .(1)

where, $r=$ distance between two charge

Now, 25% charge is transferred from one another,

$q_1=q-\frac{q}{4}=\frac{3q}{4}$

$q_2=-q+\frac{q}{4}=\frac{3q}{4}$

The force acting is given by

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}.\frac{q_1.q_2}{r^2}$

$F^{\prime }=\frac{9}{16}.\frac{q^2}{4\pi {\epsilon }_0{r}^2}$

$F^{\prime }=\frac{9}{16}F$ (from equation 1)

The correct option is D.