On increasing plate separation of a charged condenser, energy

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2. Electric Potential and Capacitance

easy

On increasing the plate separation of a charged condenser, the energy

Increases

Decreases

Remains unchanged

Becomes zero

Solution

(a) Energy $U = \frac{1}{2}\frac{{{Q^2}}}{C}$ for a charged capacitor charge $Q$ is constant and with the increase in separation $C$ will decrease $\left( {C \propto \frac{1}{d}} \right)$, So overall $U$ will increase.

Standard 12

Physics

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

medium

(IIT-1995)

A capacitor $4\,\mu F$ charged to $50\, V$ is connected to another capacitor of $2\,\mu F$ charged to $100 \,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $({10^{ – 2}}\,J)$ is

hard

If charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of electric field between the plates is $E$ then force on each plate of a parallel plate capacitor will be

hard

Charge $'q'$ on a capacitor varies with voltage $'V'$ as shown. The area of $\Delta OPM$ represents

easy

A capacitor $C$ is charged to a potential difference $V$ and battery is disconnected. Now if the capacitor plates are brought close slowly by some distance :

normal