Advertisements
Advertisements
Question
A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V–1. A similar capacitor with no dielectric is charged to U0 = 78V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
Advertisements
Solution
Let the final voltage be U: If C is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is
Q1 = CU
The capacitor with the dielectric has a capacitance εC. Hence the charge on the capacitor is
Q2 = εU = α CU2
The initial charge on the capacitor that was charged is
Q0 = CU0
From the conservation of charges,
Q0 = Q1 + Q2
Or, CU0 = CU + α CU2
⇒ αU2 + U – u0 = 0
∴ U = `(-1 +- sqrt(1 + 4αU_0))/(2α)`
= `(-1 +- sqrt(1 + 624))/4`
= `(-1 +- sqrt(625))/4` volts
As U is positive
U = `(sqrt(625) - 1)/4 = 24/4` = 6V
APPEARS IN
RELATED QUESTIONS
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.
A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm and of radii 2 mm and 4 mm. (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.
A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge of the capacitor to become 12.6 μC. Find the resistance of the circuit.
A capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure . Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.
A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?
A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constant K1 and K2 are filled in the gap as shown in figure . Find the capacitance.
A sphercial capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure . Calculate the capacitance.
You are provided with 8 μF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 μF.
Obtain the expression for capacitance for a parallel plate capacitor.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1), lightning will occur.
- If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
- In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be at ______.
The capacitance of a parallel plate capacitor is 60 µF. If the distance between the plates is tripled and area doubled then new capacitance will be ______.
A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be ______.
The displacement current of 4.425 µA is developed in the space between the plates of the parallel plate capacitor when voltage is changing at a rate of 106 Vs-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x × 10-3 m. The value of x is ______.
(Permittivity of free space, ε0 = 8.85 × 10-12C2N-1m-2).
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor.
The thickness of the dielectric slab is `3/4`d, where 'd' is the separation between the plate of the parallel plate capacitor.
The new capacitance (C') in terms of the original capacitance (C0) is given by the following relation:
Two plates A and B of a parallel plate capacitor are arranged in such a way, that the area of each plate is S = 5 × 10-3 m 2 and distance between them is d = 8.85 mm. Plate A has a positive charge q1 = 10-10 C and Plate B has charge q2 = + 2 × 10-10 C. Then the charge induced on the plate B due to the plate A be - (....... × 10-11 )C
A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness `3/8 d and d/2`, respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K, the value of K1 is:
