Question

Three identical spheres, each having charge q and radius R, are kept in such a way that each touches the other two. The magnitude of electric force on any sphere due to other two is?

Answer 1

$\textsf{Hello dear!}$

$\text{ \bullet For external points a charged sphere behaves as if the whole of its charge}\\~~~~\text{is concentrated at its center.}$

Force of A due to B:

$F_{AB} = \dfrac{1}{4\pi \mathcal{E}_0}.\dfrac{q^2} \left(2R\right)^2} =\dfrac{1}{4\pi \mathcal{E}_0}.\dfrac{q^2}{4R^2}\textsf{ along BA}$

Force on A due to C:

$F_{AC} = \dfrac{1}{4\pi \mathcal{E}_0}.\dfrac{q^2}{\left(2R\right)^2} =\dfrac{1}{4\pi \mathcal{E}_0}.\dfrac{q^2}{4R^2}\textsf{ along CA}$

$\because \textsf{the angle between BA and CA is 60}^o \ and,\\ \sf |F_{AB} | = |F_{AC}| = F$

$\therefore \sf F_A = \sqrt{F^2 + F^2 + 2F.F.\cos 60} = \sqrt{3F}$

$\sf \boxed{F_A = \dfrac{1}{4\pi \mathcal{E}_0}\dfrac{\sqrt3}{4}\left(\dfrac{q}{R}\right)^2}$

Answer 2

Solution:

Given:

Three identical spheres, each having charge q and radius R, are kept in such a way that each touches the other two.

Find:

The magnitude of electric force on any sphere due to other two is?

Step-by-step:

Force of A => B

Fab = 1/4π + εo° * q²/(2R) ² = 1/4πεo° q²/4R² = BA

Force of A => C

Eac = 1/4πεo° * q²/(2R) ² = 1/4πεo° q²/4R² = CA

BA and CA = 60°

FA = √F² + F² + F² * F * cos 60° = √3F

FA = 1/πεo √3/4 (q/R)²

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