Subjective Type

A parallel-plate capacitor with the plate area $$100 \,cm^2$$ and the separation between the plats $$1.0 \, cm$$ is connected across a battery of emf $$24$$ volts. Find the force of attraction between the plates.

Solution

The surface area of the capacitor plates $$A=100cm^2=10^{}m^2$$
The separation between the plates is $$d=1cm=10^{-2}m$$
The emf of the battery is $$V=24V_0$$

$$\therefore$$ The capacitance od the capacitor is given as:
$$C=\dfrac {\epsilon_0A }{ d }$$

$$=\dfrac { 8.85\times 10^{-12}\times 10^{-2} }{10^{-2} }$$

$$=8.85\times 10^{-12}$$

$$\therefore$$ The energy stored in the capacitor is given as:
$$E=(1/2)CV^2$$

$$=(1/2)\times 10^{-12}\times (24)^2=2548.8\times 10^{-12}$$

$$\therefore$$ The forced attraction between the plates $$=\dfrac { E}{ d} =\dfrac {2548.8\times 10^{-12} }{10^{-2} }= 2.54\times 10^{-7}N$$

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