Advertisements
Advertisements
Question
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Advertisements
Solution
Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 0.5 × 10−2 m
Capacitance of a parallel plate capacitor is given by the relation,
C = `(in_0"A")/"d"`
A = `("Cd")/in_0`
Where,
`in_0` = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2
∴ A = `(2 xx 0.5 xx 10^-2)/(8.85 xx 10^-12)` = 1130 km2
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.
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.
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.
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 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 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.
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.
In a parallel plate capacitor, the capacity increases if ______.
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.
What is the capacitance of a parallel plate capacitor with a dielectric medium of dielectric constant \[K\]?
Substituting \[V=\frac{Qd}{\varepsilon_0 A}\] into \[C=\frac{Q}{V}\] gives which capacitance?
What is the effect of increasing plate area \[A\] on capacitance, and why?
What is the effect of increasing plate separation \[d\] on capacitance?
Where is electrical energy stored in a parallel plate capacitor, and how can it be used?
For the same voltage, what change allows more charge to be stored and more energy to be stored in the capacitor?
In electronic circuits, what are filters and oscillators used for?
How are capacitors used for energy storage in power supplies?
