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# The electrostatic force on a small sphere of charge 0.4 x10^-6 C due to another small sphere of charge -0.8x10^-6 C in air is -0.2 N. (a) What is the distance between the two spheres?

Q 1.2: The electrostatic force on a small sphere of charge $0.4 \mu C$ due to another small sphere of charge $-0.8 \mu C$ in air is $-0.2N$. (a) What is the distance between the two spheres?

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Given,

$q_{1}$ = $0.4 \mu C$

$q_{2}$ =  $-0.8 \mu C$

F =  $-0.2N$ (Attractive)

We know,

Force between two charged particles, $q_{1}$  and $q_{2}$ separated by a distance r.

$F = \frac{1}{4\pi \epsilon _{0}} \frac{q_{1}q_{2}}{r^2}$

$\\ -0.2 = 9\times10^9 \times \frac{(0.4\times10^{-6})(-0.8\times10^{-6})}{r^2} \\ \implies r^2 = 9\times10^9\times0.32\times10^{-12}/0.2 \\ \implies r^2 = 9\times0.16\times10^{-2} \\ \implies r = 1.2\times10^{-1} m = 0.12 m =12 cm$

Therefore, the distance between the two charged spheres is 12 cm.

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