Advertisements
Advertisements
Question
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?
Advertisements
Solution
Given: A = 6 × 10−3 m2
V = 100 volts
d = 3 mm = 3 × 10−3 m
To find: Capacitance C = ?
Charge on each strip = ?
Formula: Capacitance (C) = `(ε_0 A)/d`
= `(8.854 xx 10^-12 xx 6 xx 10^-3)/(3 xx 10^-3)`
= `(53.124 xx 10^-15)/(3 xx 10^-3)`
= 17.7 × 10−12
= 1.77 × 10−11
Charge on a capacitor q = C × V
= 1.77 × 10−11 × 100
= 1.77 × 10−9 C
= 1.77 nC
∴ Charge on one strip = + 1.77 × 10−9 C
Charge on the second strip = −1.77 × 10−9 C
RELATED QUESTIONS
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
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.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
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 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 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.
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.
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
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.
When a dielectric slab of dielectric constant \[K\] is inserted between the plates, what is the electric field \[E'\] due to polarisation of the dielectric?
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?
