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Standard XII

Physics

Capacitance of a Random Conductor

Calculate the...

Question

Calculate the capacitance (in farads) of a charged, parallel-plate capacitor. Given the potential difference between the plates is equal to $X$ volts and the amount of charge on the POSITIVE plate is equal to $Y$ coulombs.

\frac{X}{2 Y}

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\frac{Y}{2 X}

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\frac{Y}{X}

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\frac{2 Y}{X}

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\frac{2 X}{Y}

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Solution

The correct option is C $\frac{Y}{X}$
We know that for a capacitor, the magnitude of charge $Q = C V$
Here, $Q = Y$ coulomb and $V = X$ volts
Thus, the capacitance $C = Q / V = Y / X$

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Calculate the capacitance (in farads) of a charged, parallel-plate capacitor. Given the potential difference between the plates is equal to X volts and the amount of charge on the POSITIVE plate is equal to Y coulombs.

The correct option is C YXWe know that for a capacitor, the magnitude of charge Q=CVHere, Q=Y coulomb and V=X voltsThus, the capacitance C=Q/V=Y/X

Calculate the capacitance (in farads) of a charged, parallel-plate capacitor. Given the potential difference between the plates is equal to $X$ volts and the amount of charge on the POSITIVE plate is equal to $Y$ coulombs.

The correct option is C $\frac{Y}{X}$
We know that for a capacitor, the magnitude of charge $Q = C V$
Here, $Q = Y$ coulomb and $V = X$ volts
Thus, the capacitance $C = Q / V = Y / X$