Q158 / 180 · KCET 2008 · Physics

Two identical metal spheres charged with $+12 \mu \mathrm{F}$ and $-8 \mu \mathrm{F}$ are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is

  • A
    $12: 1$
  • B
    $8: 1$
  • Correct answer
    C
    $24: 1$
  • D
    $4: 1$

Solution

$\mathrm{F}_{\text {initial }}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{12 \times(-8)}{\mathrm{r}^{2}} ; \mathrm{F}_{\text {initial }}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{96}{\mathrm{r}^{2}}$ where is the distance between them. When the charges are brought in contact, then $\mathrm{q}_{1}=\mathrm{q}_{2}=\frac{12-8}{2}=\frac{4}{2}=2 \mu \mathrm{F}$ $\therefore \quad \mathrm{F}_{\text {final }}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{2 \times 2}{\mathrm{r}^{2}}=\frac{4}{\mathrm{r}^{2}} \times \frac{1}{4 \pi \varepsilon_{0}}$ $\Rightarrow \quad|\mathrm{F}|_{\text {final }}=\frac{4}{\mathrm{r}^{2}} \times \frac{1}{4 \pi \varepsilon_{0}}$ $\frac{|\mathrm{F}|_{\text {initial }}}{|\mathrm{F}|_{\text {final }}}=\frac{96}{4}=24$

Hence, the correct answer is option: 'C'.

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