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Question
Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
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Solution
Consider a parallel plate capacitor which is connected across a battery. As soon as the charges from the battery reach one plate, due to insulating gap charge is not able to move further to the other plate. Thus, the positive charge is developed at one plate and a negative charge is developed on the other. As the amount of charge increases on the plates, a voltage is developed across the capacitor that is opposite to the applied voltage. Hence, the current flowing in the circuit decreases and gradually becomes zero. Thus, the charge is developed on the capacitor.
Energy Stored in a Charged Capacitor
The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.
Let us assume that initially, both the plates are uncharged. Now, we have to repeatedly remove small positive charges from one plate and transfer them to the other plate.
Let
q → Total quantity of charge transferred
V → Potential difference between the two plates
Then
q = CV
Now, when an additional small charge dq is transferred from the negative plate to the positive plate, the small work done is given by,
`dW = Vdq = q/C dq`
The total work is done in transferring charge Q is given by,
`W = ∫_0^Q q/C dq = 1/C∫_0^Q qdq = 1/C[q^2/2]_0^Q`
`W = Q^2/(2C)`
This work is stored as electrostatic potential energy U in the capacitor.
`U = Q^2/(2C)`
`U = (CV)^2/(2C) ...[∵ Q = CV]`
`U = 1/2 CV^2`
RELATED QUESTIONS
The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
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A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:
- Key K is kept closed and plates of capacitors are moved apart using insulating handle.
- Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
- In A: Q remains same but C changes.
- In B: V remains same but C changes.
- In A: V remains same and hence Q changes.
- In B: Q remains same and hence V changes.
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For a large sheet of charge, what is the electric field due to each plate in terms of surface charge density \[\sigma\]?
What is the effect of increasing plate area \[A\] on capacitance, and why?
What is the effect of inserting a dielectric with \[K>1\] between the plates?
When a dielectric slab of dielectric constant \[K\] is inserted between the plates, what is the electric field \[E'\] due to polarisation of the dielectric?
