Two metal pieces having a potential difference of 800 V are 0.02 m apart horizontally. A particle of

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2. Electric Potential and Capacitance

medium

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

$e$

$3e$

$6e$

$8e$

Solution

સંતુલન માટે $QE = Mg ⇒ (Ne)E = Mg$

$N\,\, = \,\,\frac{{Mg}}{{eE}}\,\,\,\left\{ {\left. {E\,\, = \,\,\frac{V}{d}} \right\}} \right.\,\,$

$ = \,\,\frac{{1.96\,\, \times \,\,{{10}^{ – 15}}\, \times \,\,10\,}}{{1.6\,\, \times \,\,{{10}^{ – 19}}\,\, \times \,\,\left( {800/0.02} \right)}}\,\, = \,\,3\,$

$\,\therefore \,\,Q\,\, = \,\,Ne\,\, = \,\,3e$

Standard 12

Physics

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easy

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easy

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medium

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medium

(IIT-1992)

The potential function of an electrostatic field is given by $V = 2x^2$. Determine the electric field strength at the point $(2\,m, 0, 3\,m)$

easy

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

Solution

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