Capacitance of a parallel plate capacitor is 12,μ ,F . If distance between plates is doubled and ar

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2. Electric Potential and Capacitance

easy

The capacitance of a parallel plate capacitor is $12\,\mu \,F$. If the distance between the plates is doubled and area is halved, then new capacitance will be.........$\mu \,F$

$8$

$6$

$4$

$3$

Solution

(d) $C = \frac{{{\varepsilon _0}A}}{d}$ As $A \to \frac{1}{2}$ times and $d \to 2$ times
So $C \to \frac{1}{4}$times i.e. $C' = \frac{1}{4}C = \frac{{12}}{4} = 3\,\mu F$

Standard 12

Physics

Dimension of Capacitance is

medium

(AIIMS-2019)

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The charges on two plates of a $10\,\mu f$ capacitor are $5\,\mu C$ and $15\,\mu C$ then the potential difference across the capacitor plates is……..$V$

medium

A condenser of $2\,\mu F$ capacitance is charged steadily from $0$ to $5$ $Coulomb$. Which of the following graphs correctly represents the variation of potential difference across its plates with respect to the charge on the condenser

medium

The capacitance of a parallel plate capacitor is $12\,\mu \,F$. If the distance between the plates is doubled and area is halved, then new capacitance will be.........$\mu \,F$

Solution

Similar Questions

Dimension of Capacitance is

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