Advertisements
Advertisements
Question
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
Advertisements
Solution
Given:-
Area of the plates, A = 20 cm2
Separation between the plates, d = 1 mm
Dielectric constant, k = 5
Emf of the battery, E = 6 V
Resistance of the circuit, R = 100 × 103 Ω
The capacitance of a parallel-plate capacitor,
\[C = \frac{K \in_0 A}{d}\]
\[ = \frac{5 \times 8 . 85 \times {10}^{- 12} \times 20 \times {10}^{- 4}}{1 \times {10}^{- 3}}\]
\[ = \frac{10 \times 8 . 85 \times {10}^{- 12} \times 20 \times {10}^{- 4}}{1 \times {10}^{- 3}}\]
\[ = 88 . 5 \times {10}^{- 12} C\]
After the connections are made, growth of charge through the capacitor,
\[Q = EC\left(1 − e^{- \frac{t}{RC}}\right)\]
\[=6\times88.5\times10^{-12}\left(1-e^{- \frac{8 . 9}{8 . 85}}\right)\]
= 335.6 × 10−12 C
Thus, energy stored in the capacitor,
\[U = \frac{1}{2}\frac{Q^2}{C}\]
\[ = \frac{1}{2} \times \frac{335 . 6 \times 335 . 6 \times {10}^{- 24}}{88 . 5 \times {10}^{- 12}}\]
\[ = \frac{335 . 6 \times 335 . 6}{88 . 5 \times 2} \times {10}^{- 12} \]
\[ = 6 . 3 \times {10}^{- 10} J\]
APPEARS IN
RELATED QUESTIONS
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
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 d/2, 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.
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
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
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.
Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
For a one dimensional electric field, the correct relation of E and potential V is _________.
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.
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 ______.
