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.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
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.
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 ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].
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 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.
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 of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
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 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.
For a one dimensional electric field, the correct relation of E and potential V is _________.
In a parallel plate capacitor, the capacity increases if ______.
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.
Two charges – q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.
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 ______.
