One end of 10 cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball having a mass of 10 g and a charge of $4.0 \times 10^{- 6} C .$ In equilibrium , the thread makes an angle of $60^{\circ}$ with the vertical. Find the surface charge density on the plate.

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Solution

$T c o s 60^{\circ} = m g$

$T s i n 60^{\circ} = q E$

$t a n 60^{\circ} = \frac{q E}{m g}$

$E = \frac{m g t a n 60^{\circ}}{q} = \frac{σ}{2 \in_{0}}$

So, $σ = \frac{2 \in_{0} m g t a n 60^{\circ}}{q}$

$= \frac{2 \times 8.8 \times 10^{- 12} \times 10 \times 10^{- 3} \times 9.8 \times 1.732}{4.0 \times 10^{- 6}}$

$= 7.5 \times 10^{- 7} c / m^{2}$

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T cos 60∘=mg T sin 60∘=qE tan 60∘=qEmg E=mg tan 60∘q=σ2 ∈0 So, σ=2 ∈0 mg tan 60∘q =2×8.8×10−12×10×10−3×9.8×1.7324.0×10−6 =7.5×10−7 c/m2