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प्रश्न
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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उत्तर
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`
संबंधित प्रश्न
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
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.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1/2)` QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor `1/2`.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Define the capacitance of a capacitor and its SI unit.
In a parallel plate capacitor, the capacity increases if ______.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
