Two point charges $(A$ and $B)+4 q$ and $-4 q$ are placed along a line separated by a distance ' $r$ '.…

Two point charges $(A$ and $B)+4 q$ and $-4 q$ are placed along a line separated by a distance ' $r$ '. Force acting between them is F. If $25 \%$ of charge from point A is transferred to that at point $B$, the force between the charges now becomes
  1. $\frac{3}{4} \mathrm{~F}$
  2. $\frac{4}{3} \cdot \mathrm{~F}$
  3. $\frac{9}{16} \mathrm{~F}$
  4. $\frac{16}{9} \mathrm{~F}$

Solution

Force acting between given charges $+4 q$ and $-4 q$ is, $F=\frac{-16 q^2}{4 \pi \varepsilon_0 r^2}$...(i) When $25 \%$ of charges ảre transferred, charge on point A becomes, $q_1=+4 q-0.25(+4 q)=+3 q$
Charge on point B becomes, $q_2=-4 q+0.25(+4 q)=-3 q$ $\therefore \quad$ The new force between points $A$ and $B$ will be, $\mathrm{F}^{\prime}=\frac{(3 \mathrm{q}) \times(-3 \mathrm{q})}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{-9 \mathrm{q}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}$ Multiplying and dividing by 16 , $F^{\prime}=\frac{9}{16} \times\left(\frac{-16 q^2}{4 \pi \varepsilon_0 r^2}\right)=\frac{9}{16} F \ldots[\operatorname{From}(i)]$

Asked in: MHT CET 2024 (10 May Shift 2)

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