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Question

સંતુલન માટે $QE = Mg &rArr; (Ne)E = Mg$

$N\,\, = \,\,\frac{{Mg}}{{eE}}\,\,\,\left\{ {\left. {E\,\, = \,\,\frac{V}{d}} \right\}} \right.\,\,$

Vedclass · Accepted Answer

$3e$

Vedclass · Answer

સંતુલન માટે $QE = Mg &rArr; (Ne)E = Mg$

$N\,\, = \,\,\frac{{Mg}}{{eE}}\,\,\,\left\{ {\left. {E\,\, = \,\,\frac{V}{d}} ight\}} ight.\,\,$

$ = \,\,\frac{{1.96\,\, 	imes \,\,{{10}^{ - 15}}\, 	imes \,\,10\,}}{{1.6\,\, 	imes \,\,{{10}^{ - 19}}\,\, 	imes \,\,\left( {800/0.02} ight)}}\,\, = \,\,3\,$

$\,	herefore \,\,Q\,\, = \,\,Ne\,\, = \,\,3e$

Two metal pieces having a potential difference of $800 \;V$ are $0.02\; m$ apart horizontally. A particle of mass $1.96 \times 10^{-15} \;kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge, then charge on the particle is

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