CUET UG Physics Booster Test – 3 Capacitance and Energy Storage

A capacitor is fundamentally a system of two conductors separated by an insulating medium. A single isolated conductor may also be treated as a capacitor by assuming the second conductor is at infinity. Therefore, 1 → c. When connected to a battery, charge is transferred such that one conductor acquires charge and the other acquires charge , giving 2 → a. The conductor plates are separated by an insulating medium, so 3 → d. The intervening medium physically separates the conductors and therefore 4 → b. These concepts form the basis of capacitor construction and operation discussed in NCERT.

Capacitance is defined as For a given capacitor, capacitance depends only on its geometry and dielectric medium and not on the actual values of charge or potential difference. Thus statement 2 is correct. Since charge and potential difference are directly proportional, making statement 4 correct. A large capacitance means a large amount of charge can be stored for a relatively small potential difference, so statement 3 is correct. Statement 1 is incorrect because doubling the charge also doubles the potential difference, leaving capacitance unchanged.

The capacitance of a conductor system depends upon its geometrical configuration and the dielectric medium present. Important factors include the size of the conductors, their shape, and the distance separating them. Therefore statements 2, 3 and 4 are correct. Statement 1 is incorrect because capacitance does not depend on the amount of charge stored. The same capacitor retains its capacitance whether it stores a small charge or a large charge.

When a dielectric is placed in an electric field, its molecules become polarized, producing induced dipoles. These dipoles create bound surface charges whose electric field opposes the external field. As a result, the net electric field inside the dielectric decreases. Consequently, for a fixed charge, the potential difference decreases to and the capacitance increases by a factor . Therefore statement B is incorrect because the induced field does not enhance the external field; it opposes it.

Dielectric strength is defined as the maximum electric field a dielectric can withstand before electrical breakdown occurs. Therefore statement 1 is correct. Breakdown causes ionization and charge leakage, thereby limiting the amount of charge that can be safely stored in a capacitor, making statement 2 correct. For air, the dielectric strength is approximately thus statement 3 is correct. Statement 4 is incorrect because exceeding dielectric strength does not make capacitance instantly zero; instead, it causes dielectric breakdown and leakage current.

Every dielectric medium has a maximum electric field that it can withstand without breakdown. When the electric field exceeds this value, the surrounding air becomes ionized. The molecules of air split into positive ions and free electrons. These charged particles are accelerated by the strong electric field toward oppositely charged plates. As a result, charge begins to leak from one plate to the other, gradually neutralizing the charges stored on the capacitor plates. This process is called dielectric breakdown. In air, the dielectric strength is approximately . Once breakdown begins, the insulating property of the medium is lost and the capacitor can no longer effectively store charge.

For a parallel plate capacitor, the positively charged plate produces an electric field of magnitude on both sides of the plate. Similarly, the negatively charged plate also produces a field of magnitude on both sides. In the outer region above the positive plate, these two fields are equal in magnitude but opposite in direction. Therefore, they cancel completely. Hence, The field exists only in the region between the plates where the fields reinforce each other. This is the basis for the localized electric field approximation used in the derivation of the parallel plate capacitor formula.

For an ideal parallel plate capacitor with plate separation much smaller than plate dimensions, the electric field between the plates is approximately uniform. The potential difference between the plates is obtained by integrating the electric field along the direction of separation: Since is constant, This expression is fundamental in deriving the capacitance formula The derivation assumes negligible fringing so that the electric field remains uniform throughout the region between the plates.

For a parallel plate capacitor in vacuum, Given Therefore, This area is extraordinarily large. A square plate of this area would have a side length of approximately or about 30 km. This calculation shows why ordinary capacitors require either extremely large surface areas, very small separations, or dielectric materials to achieve large capacitances.

When a dielectric slab is completely inserted into a capacitor, polarization occurs and bound surface charges are induced. The net electric field inside the dielectric becomes therefore . The potential difference becomes thus . If the capacitor is disconnected from the battery, no charge can enter or leave the plates, so the free charge remains unchanged, giving . The capacitance becomes hence . Thus the correct matching is:

The dielectric constant is defined as the ratio of the permittivity of a medium to the permittivity of free space: Hence is dimensionless, making statement 1 correct. Rearranging gives therefore statement 2 is correct. For vacuum, so , making statement 3 correct. Statement 4 is incorrect because ordinary dielectric materials have dielectric constants greater than or equal to 1. Conductors are not assigned dielectric constants in this context.

The dielectric constant is defined as where is the capacitance with dielectric and is the vacuum capacitance. Thus statement 3 is correct. Rearranging gives which serves as a general definition of dielectric constant, making statement 1 correct. Since for ordinary dielectrics, insertion of a dielectric increases capacitance, making statement 2 correct. Statement 4 is incorrect because the ratio definition of dielectric constant is applicable to all capacitor geometries.

In a series combination, the same charge appears on each capacitor because the intermediate conductors remain electrically neutral. Therefore each capacitor carries charges and , making statement 1 correct. The total applied voltage equals the sum of individual voltage drops: thus statement 2 is correct. Statement 3 is incorrect because the equivalent capacitance of capacitors in series is always smaller than the smallest individual capacitance. Statement 4 is incorrect because each plate acquires charge; only the net charge of the connecting conductor is zero.

For capacitors connected in series, and Hence statements A and B are correct. The equivalent capacitance is always less than the smallest capacitor in the combination, making statement C correct. In a series arrangement, each capacitor carries the same charge magnitude . Therefore the charge on the equivalent capacitor is not ; instead, Thus statement D is incorrect.

In a parallel combination, every capacitor experiences the same potential difference . Using the capacitor relation for each capacitor, and Therefore option B is correct. Capacitors with larger capacitance store proportionally larger charges when connected across the same voltage source. This principle also leads to the total charge relation and the effective capacitance relation

The three capacitors , , and are connected in series. This series combination is connected in parallel with . For capacitors in parallel, Therefore, the equivalent capacitance of the complete network is approximately . This question combines both series and parallel capacitor rules and is a standard NCERT application problem.

While charging a capacitor, charge is transferred gradually from one plate to the other. At an intermediate stage, when the capacitor has charge , the potential difference is The small work done in transferring an additional charge is Substituting , Integrating from 0 to , This work is stored as electrostatic potential energy in the capacitor.

A charged capacitor stores energy in its electric field. The stored energy is given by or When the capacitor is connected through a conducting path, charge flows from one plate to the other and the stored electric field collapses. During this process, the electrostatic energy is released and may appear as heat in a resistor, light in a bulb, or useful electrical work in a circuit. Thus the stored potential energy does not remain trapped in the capacitor but is released during discharge.

The electric energy density in vacuum is If the electric field is doubled, Substituting into the formula, Thus, when the electric field doubles, the energy density becomes four times its original value.

Energy density is defined as the energy stored per unit volume: For a parallel plate capacitor, the electric field occupies the volume where is the plate area and is the plate separation. Therefore, Rearranging, This expression shows that the total electrostatic energy stored in a capacitor equals the energy density multiplied by the volume occupied by the electric field. The same interpretation leads to the general energy density formula which is valid for any electrostatic field configuration.