Advertisements
Advertisements
Question
A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
Advertisements
Solution
(i)
Q = CV
`Q = ((epsi_0A)/d) (Ed)`
`Q = epsi_0AE`
`therefore E =Q/(epsi_0A)`
Therefore, the electric field between the parallel plates depends only on the charge and the plate area. It does not depend on the distance between the plates.
Since the charge as well as the area of the plates does not change, the electric field between the plates also does not change.
(ii)
Let the initial capacitance be C and the final capacitance be C'.
Accordingly,
`C = (epsiA)/d`
`C' = (epsi_0A)/(2d)`
`C/C' = 2`
`C' \ C/2`
Hence, the capacitance of the capacitor gets halved when the distance between the plates is doubled.
(iii)
Energy of a capacitor, U `=1/2 (Q_2)/C`
Since Q remains the same but the capacitance decreases,
`U' = 1/2 (Q^2)/((C/2))`
`U/U' = 1/2`
U' = 2U
The energy stored in the capacitor gets doubled when the distance between the plates is doubled.
APPEARS IN
RELATED QUESTIONS
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3m2 and the separation between the plates is 3 mm.
- Calculate the capacitance of the capacitor.
- If this capacitor is connected to 100 V supply, what would be the charge on each plate?
- How would charge on the plates be affected, if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.
A parallel-plate capacitor is filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will ______.
What is a capacitor that consists of two large, parallel, conducting plates separated by a small distance called?
For two plates separated by distance \[d\], what is the capacitance of a parallel plate capacitor without a dielectric medium?
What is the effect of increasing plate area \[A\] on capacitance, and why?
Which description gives the working principle of a parallel plate capacitor?
Where is electrical energy stored in a parallel plate capacitor, and how can it be used?
In electronic circuits, what are filters and oscillators used for?
