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# One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged non-conducting plate and the other end is fastened to a small ball of mass 10 g and a charge of $4·0×{10}^{-6}\mathrm{C}$. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.

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Solution

## There are two forces acting on the ball. These are (1) Weight of the ball, W = mg (2) Coulomb force acting on the charged ball due to the electric field of the plate, F = qE Due to these forces,a tension develops in the thread. Let the surface charge density on the plate be σ. Electric field of a plate, E = $\frac{\sigma }{2{\in }_{0}}$ It is given that in equilibrium, the thread makes an angle of 60° with the vertical. Resolving the tension in the string along horizontal and vertical directions, we get: $T\mathrm{cos}60°=mg\phantom{\rule{0ex}{0ex}}T\mathrm{sin}60°=\mathit{}qE\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}60°=\frac{\mathit{q}\mathit{E}}{\mathit{m}\mathit{g}}\phantom{\rule{0ex}{0ex}}⇒E=\frac{m\mathrm{g}\mathrm{tan}60°}{q}$ Also, electric field due to a plate, $E=\frac{\sigma }{2{\in }_{0}}=\frac{mg\mathrm{tan}{60}^{\mathrm{o}}}{q}\phantom{\rule{0ex}{0ex}}\sigma =\frac{2{\in }_{0}m\mathrm{g}\mathrm{tan}60°}{q}\phantom{\rule{0ex}{0ex}}\sigma =\frac{2×\left(8.85×{10}^{-12}\right)×\left(10×{10}^{-3}×9.8\right)×1.732}{4.0×{10}^{-6}}\phantom{\rule{0ex}{0ex}}\sigma =7.5×{10}^{-7}\mathrm{C}/{\mathrm{m}}^{2}$  Suggest Corrections  0      Related Videos   Magnetic Force
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