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Byju's Answer

Standard XII

Physics

Partially Filled Dielectrics

Two conductin...

Question

Two conducting plates of parallel plate capacitor are placed at a small distance from each other as shown in figure. a, b, c and d are charges on different surfaces.

Then, potential difference between plates ( $V_{H} - V_{L}$ ) is

\frac{a}{capacitance}

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\frac{b}{capacitance}

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\frac{c}{capacitance}

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\frac{d}{capacitance}

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Solution

The correct option is B $\frac{b}{capacitance}$
$Δ V = \frac{q_{i n n e r p l a t e (+ v e)}}{C} = \frac{b}{capacitance}$
or
$Δ V = - \frac{q_{i n n e r p l a t e (- v e)}}{C} = \frac{- c}{capacitance}$

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Two conducting plates of parallel plate capacitor are placed at a small distance from each other as shown in figure. a, b, c and d are charges on different surfaces. Then, potential difference between plates (VH−VL ) is

The correct option is B bcapacitanceΔV=qinner plate(+ve)C=bcapacitance or ΔV=−qinner plate(−ve)C=−ccapacitance

Two conducting plates of parallel plate capacitor are placed at a small distance from each other as shown in figure. a, b, c and d are charges on different surfaces.

Then, potential difference between plates ( $V_{H} - V_{L}$ ) is

The correct option is B $\frac{b}{capacitance}$
$Δ V = \frac{q_{i n n e r p l a t e (+ v e)}}{C} = \frac{b}{capacitance}$
or
$Δ V = - \frac{q_{i n n e r p l a t e (- v e)}}{C} = \frac{- c}{capacitance}$