CUET UG Physics Booster Test – 2 Capacitance and Energy Storage

A capacitor consists of two conductors separated by an insulating medium. When connected to a source, one conductor acquires a charge while the other acquires an equal and opposite charge . Therefore, the algebraic sum of charges on the capacitor is zero, not . Statement A correctly describes the structure of a capacitor. Statement B is correct because equal and opposite charges appear on the conductors. Statement C is also correct since the capacitance of an isolated conductor can be discussed by assuming the second conductor to be at infinity. Hence statement D is the incorrect statement because it ignores the opposite sign of charges.

The capacitance of a capacitor is given by Therefore, Substituting the given values, Thus the potential difference across the capacitor plates is 10 V. This result follows directly from the definition of capacitance. The unit verification is which gives volt.

For a parallel plate capacitor, An increase in plate area increases capacitance, therefore . An increase in plate separation decreases capacitance, therefore . Since a lower potential for a given charge implies a larger capacitance, therefore . Capacitance itself depends on geometry and dielectric medium, therefore . Hence the correct matching is

When a dielectric is inserted between capacitor plates, polarization occurs. The induced dipoles create an electric field opposite to the applied field. This opposing field reduces the net electric field inside the dielectric but does not completely cancel it. Therefore statement 3 is correct and statement 1 is incorrect. Since the electric field decreases, the potential difference between the plates also decreases for the same stored charge, making statement 4 correct. The capacitance becomes and therefore increases by a factor , not decreases. Hence statement 2 is incorrect.

Dielectric strength is defined as the maximum electric field that a dielectric can withstand without electrical breakdown. Therefore statement 1 is correct. Air has a dielectric strength of approximately making statement 3 correct. Since electric field and potential difference are related by a large potential difference generally corresponds to a strong electric field, so statement 4 is correct. Statement 2 is incorrect because a larger capacitance allows more charge to be stored for the same voltage. Therefore high-capacitance systems are desirable for storing large amounts of charge.

When the electric field around a capacitor becomes very strong, it can ionise the surrounding air molecules. This produces free electrons and ions. These charged particles are accelerated by the electric field toward oppositely charged capacitor plates. As they reach the plates, they neutralise part of the stored charge, causing charge leakage and reducing the effectiveness of the capacitor. Thus statements 1, 2 and 3 are correct. Statement 4 is incorrect because charge leakage reduces the insulating effectiveness of the medium rather than improving it. This phenomenon is responsible for dielectric breakdown and limits the maximum voltage that can be applied across a capacitor.

A large charged conducting plate produces an electric field on either side of the sheet. In a parallel plate capacitor, the positively charged plate and negatively charged plate create electric fields in the same direction within the space between them. Therefore the fields add: Outside the plates, the fields are equal and opposite and therefore cancel. This result is one of the key assumptions used in deriving the capacitance of a parallel plate capacitor.

The derivation of the parallel plate capacitor formula assumes infinitely large plates so that the electric field remains perfectly uniform. In reality, capacitor plates have finite dimensions. Near the edges, electric field lines are no longer perfectly parallel and begin to curve outward. This bending of field lines is known as fringing. Fringing causes slight deviations from the ideal uniform field assumption. However, when the plate area is much larger than the square of the separation distance , the fringing effect becomes negligible and the standard capacitor formula remains valid.

For a parallel plate capacitor in vacuum, and Since the potential difference becomes Using the definition we obtain This derivation assumes a nearly uniform electric field between the plates and negligible fringing effects. The result shows that capacitance increases with plate area and decreases with separation.

The capacitance of a parallel plate capacitor is If the plate separation is halved, When a dielectric of constant completely fills the space, Unit Verification Thus the new capacitance is 96 pF.

The dielectric constant , also called the relative permittivity of a medium, is defined as the ratio of the permittivity of the medium to the permittivity of free space. Mathematically, Rearranging, This relation is fundamental in electrostatics and explains why capacitors store more charge when a dielectric medium is inserted between their plates. Since for ordinary dielectric materials, the permittivity of the medium is greater than the permittivity of vacuum. Hence the correct answer is dielectric constant .

The dielectric constant is defined as and also where is the capacitance without dielectric. Since dielectric materials increase capacitance, the value of is generally greater than 1. Therefore statement C is incorrect. Statements A and B are standard definitions given in NCERT. Statement D is also correct because inserting a dielectric multiplies the capacitance of any capacitor by the factor , provided the dielectric completely fills the region.

When capacitors are connected in series, charge flows through the circuit until electrostatic equilibrium is established. The connecting conductor between adjacent capacitors must remain field-free in electrostatic equilibrium. As a result, equal and opposite charges accumulate on the facing plates, ensuring that each capacitor carries the same magnitude of charge . The total potential difference across the combination equals the sum of the individual potential differences, not just that of the first capacitor. Capacitors do not generate charge internally; they merely store charge supplied by the battery. Therefore statement D correctly describes the equilibrium condition in a series combination.

For capacitors connected in series, Given, Therefore, Hence, The supply voltage does not affect the value of equivalent capacitance. Thus the effective capacitance of the combination is .

In a parallel combination, every capacitor is connected across the same pair of terminals. Therefore the potential difference across each capacitor is the same, making statement 1 correct. The total charge supplied by the source equals the sum of charges stored on all capacitors: Hence statement 4 is correct. Using , which gives Therefore statement 2 is also correct. Statement 3 is incorrect because the effective capacitance of a parallel combination is greater than any individual capacitance.

In a parallel combination of capacitors, all capacitors are connected across the same two terminals. Therefore, the potential difference across each capacitor is the same, making statement 4 correct. The total charge supplied by the source is distributed among the capacitors, and hence making statement 3 correct. Using the relation , which gives Therefore statement 2 is correct. Statement 1 is incorrect because the charges on individual capacitors depend on their capacitances. Capacitors with different capacitances store different amounts of charge even though the voltage is the same.

During charging, an infinitesimal charge is transferred against the potential difference . Therefore, giving . The total work done in charging a capacitor becomes thus . Initially, both conductors are uncharged, so . Since electrostatic force is conservative, the final stored energy is independent of the path followed during charging, giving . Hence the correct matching is:

The energy stored in a capacitor is not confined to the conducting plates themselves. Modern field theory interprets the stored energy as residing in the electric field established between the plates. This viewpoint leads to the concept of electric energy density, which gives the energy stored per unit volume of space. Thus the energy associated with a charged capacitor can be viewed as distributed throughout the region occupied by the electric field. Therefore statement C is correct.

The electric energy density in vacuum is thus statement 1 is correct. It is obtained by dividing the total stored electrostatic energy by the volume occupied by the electric field: therefore statement 2 is correct. Energy density represents energy stored per unit volume of space, making statement 4 correct. Statement 3 is incorrect because energy density is directly proportional to , not inversely proportional.

The expression is often derived using a parallel plate capacitor because the electric field is uniform and easy to analyze mathematically. However, the final result is not restricted to capacitors. The formula represents the energy stored per unit volume in an electric field and therefore applies to fields produced by point charges, dipoles, continuous charge distributions, and all other electrostatic configurations. Thus the result is universally valid for electrostatic fields.