Advertisements
Advertisements
Question
Obtain the expression for energy stored in the parallel plate capacitor.
Advertisements
Solution
- Capacitor not only stores the charge but also stores energy.
- When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate.
- To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
- To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = V dQ ………….(1)
where V = `"Q"/"C"`
The total work done to charge a capacitor is
W = `int_0^"Q" "Q"/"C" "dQ" = "Q"^2/"2C"` .....(2)
This work done is stored as electrostatic potential energy (UB) in the capacitor.
`"U"_"E" = "Q"^2/"2C" = 1/2 "CV"^2` ....(3)
where Q = CV is used. - This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.
`"U"_"E" = 1/2 ((ε_0"A")/"d")("Ed")^2 = 1/2 ε_0 ("Ad")"E"^2` ....(4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined `"U"_"E" = 1/2ε_0 "E"^2` .....(5) - Energy is stored in the electric field existing between the plates of the capacitor. 0nce the capacitor is allowed to discharge, the energy is retrieved.
- The energy density depends only on the electric field and not on the size of the plates of the capacitor.
APPEARS IN
RELATED QUESTIONS
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
A 5⋅0 µF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.
A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
Three circuits, each consisting of a switch 'S' and two capacitors, are initially charged, as shown in the figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor
(i) increase,
(ii) decrease, and
(iii) remains the same? Give reasons.

Derive the expression for resultant capacitance, when the capacitor is connected in parallel.
Two similar conducting spheres having charge+ q and -q are placed at 'd' seperation from each other in air. The radius of each ball is r and the separation between their centre is d (d >> r). Calculate the capacitance of the two ball system ______.

The material filled between the plates of a parallel plate capacitor has a resistivity of 200Ωm. The value of the capacitance of the capacitor is 2 pF. If a potential difference of 40V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is ______.
(given the value of relative permittivity of a material is 50.)
A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer is ______.
A capacitor of capacity 2 µF is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the potential difference across the capacitor is 6.0 V is ______ × 10-1A.
A capacitor has charge 50 µC. When the gap between the plate is filled with glass wool, then 120 µC charge flows through the battery to capacitor. The dielectric constant of glass wool is ______.
