Team Quanta gladly presents all possible short questions of Electricity & Magnetism – I’s Chapter#03: Capacitance.
Q.1 What is effect of dielectric on original electric field of capacitor?
Answer: The electric field between the plates of the capacitor reduces by the influence of dielectric by a factor equal to dielectric constant of that material.
Q.2 Name some examples in which a capacitor is used for storing charge.
Answer: Capacitors are being widely used in the electronics. It is used in radio circuits for tuning the radio. Capacitors are used in the induction motors to store charge.
Q.3 If the potential difference across a capacitor is doubled, by what factor does the energy stored change?
Answer: When the potential difference across the capacitor is doubled, the energy stored inside it increases by the factor of 4. Here is the mathematical proof:
$$U=\frac{1}{2}CV^2$$
When potential difference is doubled
$$U^\prime=\frac{1}{2}C\left(2V\right)^2$$
$$U^\prime=4\left(\frac{1}{2}CV^2\right)$$
$$U=4U$$
Q.4 Water has high dielectric constant. Why is it not used ordinarily as a dielectric material in capacitors?
Answer: Dielectric by definition is an insulator material placed between the plates of the capacitor to increase its capacitance. Water, although is not an insulator, that’s why it is not used as dielectric.
Q.5 Explain the term polarization.
Answer: The phenomenon in which negative charges appears on one face and positive charges appear on other face of the dielectric when it is placed in applied electric field is called electric polarization.
Q.6 Define the terms capacitance and capacitor. Give the principle of a capacitor.
Answer: The ability of a capacitor to store charges is called capacitance. Capacitor is a device that stores electric charges. A capacitor works on the principle that the capacitance of a conductor increases appreciably when an earthed conductor is brought near it
Q.7 What kind of energy is stored in capacitor?
Answer: The energy stored inside the capacitor is electric potential energy.
Q.8 What advantage might there be in using two identical capacitors in parallel connected in series with another identical parallel pair, rather than using a single capacitor?
Answer: In this case the potential difference would decrease between the plates of any individual capacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown.
Q.9 What happens when a non-polar molecule is placed in an electric field?
Answer: When a non-polar molecule is placed in an electric field, the center of positive charge moves in direction of field and center of negative charge in opposite direction. This separation of positive and negative charge continues till force on either of them due to external field is completely balanced by internal forces arising due to their relative displacements. The molecule develops a dipole moment known as induced dipole moment. Such a molecule is said to be polarized. The induced dipole moment disappears as soon as electric field is removed.
Q.10 Why it is dangerous to touch the terminals of a high voltage capacitor even after the applied potential difference has been turned off?
Answer: A capacitor stores energy in the electric field inside the dielectric. When extremal voltage source is removed in absence of external resistance through which the capacitor can discharge the capacitor can hold onto this energy for a very long time.
Q.11 On what factors, the capacitance of a parallel plate capacitor depends upon?
Answer: The capacitance of a parallel plate capacitor depends upon following factors:
- shape or geometry of plates
- area of plates
- distance between plates
- dielectric between plates
Q.12 A fully charged parallel-plate capacitor remains connected to a battery charge and you while slide a dielectric between the plates. What happens to its capacitance, charge, and electric field between plates?
Answer: Capacitance increases by introduction of dielectric. Charge also increases. Electric field between the plates remains constant because V = Ed and neither V nor d changes.
Q.13 What is dielectric strength of a material? In what units is it measured?
Answer: Dielectric strength of a material is defined as maximum electric field which material can withstand without breaking down. The value of dielectric strength for alternating voltages is different from that of direct voltages.
As electric field is measured in V/m, the dielectric strength is measured in kV/mm. For example, dielectric strength of air is 0.8kV/mm and that of mica is 160kV/mm.
Q.14 Suppose we want to increase the maximum operating voltage of a parallel plate capacitor. Describe how you can do this for a fixed plate separation.
Answer: This can be done by creating vacuum between plates or by placing a dielectric between plates. At very high voltages, we must cool off the plates or choose to make them of a different chemically stable material, since atoms in the plates can ionize, showing thermionic emission under high electric fields.
Q.15 The distance between plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between plates. What will be its effect on capacitance?
Answer: When metal plate is inserted between plates of a capacitor, it will act as two capacitors and the capacitance ill increase.
Q.16 Is it true that alternating current can pass through a capacitor while a direct current cannot? Explain.
Answer: It is true that the varying electric flux produced by alternating current gives rise to a displacement current through the dielectric slab. However, the direct current produces no such varying flux and fails to pass through a dielectric slab.
Q.17 Two capacitors are identical. They can be connected in series or in parallel. If we want the smallest equivalent capacitance for the combination, do you connect them in series?
Answer: When capacitors are connected in series, then equivalent capacitance is reduced.
Q.18 The sum of the charges on both plates of a capacitor is zero. What does a capacitor store?
Answer: Since both plates of capacitor have equal and opposite charge so sum of charges on both plates of capacitor is zero. A capacitor stores energy in the electric field between the plates.
Q.19 A dielectric slab is inserted between the plates of a capacitor. Describe what happened? Explain why the introduction of a dielectric changes its capacitance?
Answer: On the introduction of dielectric slab of a dielectric constant $k_e$ between plates of a capacitor, the electric field intensity at a point between the plates falls from $\sigma/\epsilon_\circ$ to $\sigma/k_e\epsilon_\circ$, where $\sigma$ is charge per unit area. This potential difference between plates decreases from $\sigma d/\epsilon_\circ$ to $\sigma d/k_e\epsilon_\circ.$. As,
$$C=\frac{q}{V}=\frac{q}{\frac{\sigma d}{k_e\epsilon_\circ}}=k_e\epsilon_\circ\frac{q}{\sigma d}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because\sigma=\frac{q}{A}$$
$$\Rightarrow C=k_e\epsilon_\circ\frac{q}{\frac{q}{A}d}=k_e\epsilon_\circ\frac{A}{d}\ \$$
Hence capacitance of capacitor increases $k_e$ times due to the introduction of dielectric slab
Q.20 For a given potential difference does a capacitor can store more r less charge with a dielectric than it does without a dielectric?
Answer: For a given potential difference V, the charge on capacitor without dielectric is and with dielectric is $q=CV$
$$q_d=C_dV$$
$$\frac{q_d}{q}=\frac{C_dV}{CV}=\frac{C_d}{C}=\frac{k_e\epsilon_\circ\ \frac{A}{d}}{\epsilon_\circ\ \ \frac{A}{d}\ }=k_e$$
A $k_e\gg1$. The capacitor will store more charge with a dielectric than it does without it.
Q.21 The capacitor is connected across a battery. How does each plate receive a charge of same magnitude even for plates of different size?
Answer: The charge on plates of capacitor is independent of size of plate. It depends on voltage given to plates. So, charge on both plates of capacitor is equal in magnitude.
Q.22 Many computer keyboard buttons are constructed of capacitors. When a key is pushed down, the soft insulator between the movable plate and the fixed plate is compressed. When the key is pressed, what happens to capacitance?
Answer: Capacitance of parallel plate capacitor is given by,
$$q=CV\ \ \ \Rightarrow q\propto V$$
When the key is pressed, the plate separation is decreased and the capacitance increases.
Q.23 Consider a parallel plate capacitor with square plates of area A and separation d in vacuum. What is effect on its capacitance when:
(a)-Distance between plates is reduced.
(b)-Area of both plates is doubled.
(c)-Potential difference between plates is doubled.
Answer: Capacitance of parallel plate capacitor is given by,
$$C=\frac{\epsilon_\circ A}{d}$$
(a)-When d is reduced, capacitance increases.
(b)-When area of each plate is doubled, capacitance becomes two times of original value.
(c)-There is no effect of potential difference on capacitance of capacitor.
Q.24 Suppose we have three capacitors and a battery. In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery? (a)-series (b)-parallel (c)-both combinations will store the same amount of energy.
Answer: When capacitors are connected in parallel, then capacitances add up and maximum energy is stored according to relation;
$$U=\frac{1}{2}CV^2$$
Q.25 Explain why the work needed to move a charge through a potential difference V is W=qv whereas the energy stored in a charged capacitor is U=qV/2 . Where does the factor 1/2 c come from?
Answer: The work done by an external agent to move a charge through a potential difference V is W =qV . To find the energy stored in a capacitor, we must add the work done to move some of charge from one plate to the other, Initially, there is no potential difference between the plates of an uncharged capacitor. As more charge is transferred from one plate to the other, the potential difference increases and average value of potential difference is V/2, so energy stored is U=qV/2.
Q.26 A capacitor stores charge q at a potential difference V. If the voltage applied by a battery to the capacitor is doubled to 2V, what happens to its capacitance?
Answer: Capacitance of parallel plates capacitor is given by,
$$C=\frac{\epsilon_\circ A}{d}$$
There is no effect of potential difference on capacitance of capacitor.
Q. 27 The plates of a capacitor are connected to a battery. What happens to the charge if the wires are removed from the battery and connected to each other?
Answer: When wires are connected to each other, charges move between the wires and the plates, until the entire conductor is at a single potential and the capacitor is discharged.
Q.28 Consider a parallel plate capacitor with square plates of area ![]()
and separation ![]()
in vacuum. What is effect on its capacitance when area of one plate is doubled?
Answer: Capacitance of parallel plate capacitor is given by,
$$C=\frac{\epsilon_\circ A}{d}$$
When area of one plate is doubled, capacitance is doubled.
Q.29 A parallel plate capacitor is connected to a battery that maintains a constant potential difference between plates. If plates are pulled away from each other, increasing their separation. What happens to charge on plates?
Answer: Capacitance of parallel plate capacitor is given by,
$$C=\frac{\epsilon_\circ A}{d}$$
By increasing $d,\ \ C$ decreases. For constant value of V, from q=CV, charge q decreases.
Q.30 Explain why a dielectric increases the maximum operating voltage of a capacitor although the physical size of the capacitor does not change.
Answer: The material of dielectric is able to support a larger electric field than air, without breaking down to pass a spark between the capacitor plates.
Q.31 The plates of a particular parallel plate capacitor are uncharged. Is the capacitance of capacitor zero?
Answer: No, capacitance of the capacitor is not zero.
Q.32 When a sheet of mica is inserted between plates of a capacitor, what happens to capacitance of capacitor?
Answer: When a sheet of mica is inserted between plates of a capacitor, it gets polarized and its electric field is opposite to original electric field. The resultant electric field between plates reduces and hence potential difference. From relation C=q/V, capacitance increases.
Q.33 Using the polar molecule description of a dielectric, explain how a dielectric affects the electric field inside a capacitor?
Answer: In polarized dielectric, each molecule aligns itself with the external electric field set up by the charged plates. The contribution of these electric dipoles pointing in the same direction reduces the total electric field. Because each dipole sets into a configuration of lower potential energy it can contribute to increasing the internal energy of the material.
Q.34 If you were asked to design a capacitor where small size and large capacitance were required, what factors would be important in your design?
Answer: Capacitance of a capacitor can be increased in following ways,
- By placing a dielectric of large dielectric constant between plates.
- By decreasing space between since capacitance in inversely proportional to separation between plates.
- By combining several capacitors in parallel.
Q.35 Show that force on each plate of a parallel plate capacitor has a magnitude qE/2, where q is charge on each plate of capacitor and E Is electric field.
Answer: Let F be force between plates, then
$$Fd=\Delta U\ \ \ \ \ \Rightarrow Fd=\frac{q^2}{2C}\ \ \ \Rightarrow Fd=\frac{q^2}{2\epsilon_{\circ\ \ }\frac{A}{d}\ }\ \ \ \ \Rightarrow Fd=\frac{qd}{2\epsilon_\circ}\left(\frac{q}{A}\right)$$
$$\Rightarrow F=\frac{q\sigma}{2\epsilon_\circ}=q\left(\frac{\sigma}{2\epsilon_\circ}\right)=qE$$
Q.36 A pair of capacitors is connected in parallel while identical pair is Connected in series. Which pair would be more dangerous to handle after being connected to the same battery?
Answer: The parallel connected capacitors are more dangerous because these store more energy due to higher equivalent capacitance.
Q.37 If the potential difference across a capacitor is doubled, by what factor does the energy stored change?
Answer: Energy stored in a capacitor is,
$$U=\frac{1}{a}qV^2$$
When V is doubled, U increases four times.
