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$$(4 \pi R^2a)P_v=(\frac{4}{3} \pi r^3)P_v$$
$$r=\sqrt{3R^2a}$$
$$0.996 \times 10^{-3}$$
$$1=\frac{Q}{4\pi E_0 \times 1 \times 10^{-2}}$$
$$Q=4 \pi \varepsilon _0 \times 10^{-2}C$$
$$\underline{Potential of Soap drop}$$
$$V=\frac{Q}{4 \pi \varepsilon _o r}$$
$$=\frac{4 \pi \varepsilon _o\times 10^{-2}}{4\pi\varepsilon _o \times 0.9966 \times 10^{-3}}$$
$$=10.03V$$