A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations
(i) key $K$ is kept closed and plates of capacitors are moved apart using insulating handle
(ii) key $K$ is opened and plates of capacitors are moved apart using insulating handle
Which of the following statements is correct?
Updated On: Jun 23, 2023
In (i), $Q$ remains same but $C$ changes
In (ii) $V$ remains same but $C$ changes
In (i) $V$ remains same and hence $Q$ changes.
In $(ii)$ both $Q$ and $V$ changes
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The Correct Option isC
Solution and Explanation
When key K is kept closed, condenser C is charged to potential V. When plates of capacitors are moved apart, its capacitance, C = $\frac {\epsilon_o A}{d}$ decreases.
As potential of condenser remains same, charge $Q = CV$ decreases.So option is correct. Once key $K$ is closed, condenser gets charged, $Q = CV$ . Now, if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains same. As plates or capacitor are moved apart, its capacity $C = \frac {epsilon_o A}{d}$ decreases.
Therefore, its potential , $V =\frac {q}{C} increases$
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Top Questions on electrostatic potential and capacitance
The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.
Some major things that we should know about electric potential:
They are denoted by V and are a scalar quantity.
It is measured in volts.
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In Parallel
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