Q: Two thin metallic spherical shells are placed concentrically with radii R1 and R2, where R1 > R2. If the shells carry charges Q1 and Q2 respectively, what is the electric potential at a point located at distance r such that R2 < r < R1? (Take k = 1/(4πϵ0))

k(Q2/r + Q1/R1). At R2 < r < R1 the point is outside the inner shell (Q2 contributes kQ2/r) and inside the outer shell (Q1 contributes the constant kQ1/R1). So V = k(Q2/r + Q1/R1).

Q: An alpha particle is made to move through a potential difference of 10⁶ V. What kinetic energy does it gain?

2 MeV. Kinetic energy gained = qV. An alpha particle has charge 2e, so KE = 2e * 10^6 V = 2x10^6 eV = 2 MeV.

Q: A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position x according to K(x) = K0 + λx, where λ is a constant. How is its capacitance C related to the vacuum capacitance C0?

C = (λd / ln(1 + λd/K0)) C0. 1/C = (1/(eps0*A)) * integral dx/(K0+lambda*x) over 0..d = (1/(eps0*A*lambda)) ln(1+lambda*d/K0). Hence C = (lambda*d/ln(1+lambda*d/K0)) * C0, with C0 = eps0*A/d.

Q: Two thin circular rings of radius R each are kept on the same axis with their centres separated by a distance d. One ring carries charge +q and the other carries charge −q. What is the potential difference between the centres of the two rings?

(q)/(2πε₀) [(1)/(R)-(1)/(√(R²+d²)) ]. V at centre of ring1 = (q/4 pi eps0)[1/R - 1/sqrt(R^2+d^2)] and centre of ring2 = its negative. Potential difference = (q/2 pi eps0)[1/R - 1/sqrt(R^2+d^2)].

Q: A uniformly charged, non-conducting circular disc of radius a lies on the floor with its symmetry axis vertical. A particle of mass m and charge q is released from rest on the axis at a height H above the disc. If it is just able to reach the disc, the electrostatic potential at the starti

(σ/2ϵ0)[(a² + H²)^(1/2) − H]. The potential on the axis of a uniformly charged disc at height z is V = (sigma/(2*eps0))*(sqrt(a^2+z^2) - z). At the starting height H, V = (sigma/(2*eps0))*(sqrt(a^2+H^2) - H).

Q: Two identical conducting plates carry positive charges Q1 and Q2, with Q2 < Q1. When they are placed close together so that they act as a parallel-plate capacitor of capacitance C, what is the potential difference between the plates?

(Q1 - Q2) / 2C. For two charged conducting plates forming a capacitor, the potential difference is V = (Q1 - Q2)/(2C).

Q: A positive charge is let go from rest in a uniform electric field. What happens to its electric potential energy?

It decreases because the charge moves in the direction of the field. A positive charge released from rest accelerates along the field toward lower potential; its potential energy decreases (converting to kinetic energy).

Q: A parallel-plate capacitor filled with air is first charged so that the potential difference across it is V volts. The charging battery is then removed, and the plate separation is increased using an insulating handle. What happens to the potential difference between the plates?

It increases. With the battery removed the charge Q is constant. Increasing separation reduces C = e0*A/d, so V = Q/C increases.

Question 1

If an electric field vector E exists in a material whose dielectric constant is K, and ε0 denotes the permittivity of free space, which expression gives the electric displacement vector?

Accepted Answer

K ε0 E. The electric displacement vector is D = epsilon*E = K*eps0*E, where K is the dielectric constant.

Question 2

If a unit positive charge is taken from point A to point B along the same equipotential surface, then what can be said about the potential difference V_A - V_B?

Accepted Answer

V_A - V_B is zero. On an equipotential surface every point is at the same potential, so V_A - V_B = 0.

Question 3

Two thin metallic spherical shells are placed concentrically with radii R1 and R2, where R1 > R2. If the shells carry charges Q1 and Q2 respectively, what is the electric potential at a point located at distance r such that R2 < r < R1? (Take k = 1/(4πϵ0))

Accepted Answer

k(Q2/r + Q1/R1). At R2 < r < R1 the point is outside the inner shell (Q2 contributes kQ2/r) and inside the outer shell (Q1 contributes the constant kQ1/R1). So V = k(Q2/r + Q1/R1).

Question 4

An alpha particle is made to move through a potential difference of 10⁶ V. What kinetic energy does it gain?

Accepted Answer

2 MeV. Kinetic energy gained = qV. An alpha particle has charge 2e, so KE = 2e * 10^6 V = 2x10^6 eV = 2 MeV.

Question 5

A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position x according to K(x) = K0 + λx, where λ is a constant. How is its capacitance C related to the vacuum capacitance C0?

Accepted Answer

C = (λd / ln(1 + λd/K0)) C0. 1/C = (1/(eps0*A)) * integral dx/(K0+lambda*x) over 0..d = (1/(eps0*A*lambda)) ln(1+lambda*d/K0). Hence C = (lambda*d/ln(1+lambda*d/K0)) * C0, with C0 = eps0*A/d.

Question 6

Two thin circular rings of radius R each are kept on the same axis with their centres separated by a distance d. One ring carries charge +q and the other carries charge −q. What is the potential difference between the centres of the two rings?

Accepted Answer

(q)/(2πε₀) [(1)/(R)-(1)/(√(R²+d²)) ]. V at centre of ring1 = (q/4 pi eps0)[1/R - 1/sqrt(R^2+d^2)] and centre of ring2 = its negative. Potential difference = (q/2 pi eps0)[1/R - 1/sqrt(R^2+d^2)].

Question 7

A uniformly charged, non-conducting circular disc of radius a lies on the floor with its symmetry axis vertical. A particle of mass m and charge q is released from rest on the axis at a height H above the disc. If it is just able to reach the disc, the electrostatic potential at the starti

Accepted Answer

(σ/2ϵ0)[(a² + H²)^(1/2) − H]. The potential on the axis of a uniformly charged disc at height z is V = (sigma/(2*eps0))*(sqrt(a^2+z^2) - z). At the starting height H, V = (sigma/(2*eps0))*(sqrt(a^2+H^2) - H).

Question 8

Two identical conducting plates carry positive charges Q1 and Q2, with Q2 < Q1. When they are placed close together so that they act as a parallel-plate capacitor of capacitance C, what is the potential difference between the plates?

Accepted Answer

(Q1 - Q2) / 2C. For two charged conducting plates forming a capacitor, the potential difference is V = (Q1 - Q2)/(2C).

Question 9

A positive charge is let go from rest in a uniform electric field. What happens to its electric potential energy?

Accepted Answer

It decreases because the charge moves in the direction of the field. A positive charge released from rest accelerates along the field toward lower potential; its potential energy decreases (converting to kinetic energy).

Question 10

A parallel-plate capacitor filled with air is first charged so that the potential difference across it is V volts. The charging battery is then removed, and the plate separation is increased using an insulating handle. What happens to the potential difference between the plates?

Accepted Answer

It increases. With the battery removed the charge Q is constant. Increasing separation reduces C = e0*A/d, so V = Q/C increases.

Question 11

A capacitor of capacitance C is initially charged and then allowed to discharge through a small coil of resistance R placed inside a thermally insulated block of mass m and specific heat capacity s. If the block’s temperature increases by ΔT, what was the potential difference V across the

Accepted Answer

√(2msΔT / C). The correct option is derived from the energy conservation principle, where the energy stored in the capacitor, given by (1/2)CV², is converted into heat energy absorbed by the block, which is calculated as msΔT. By equating these two forms of energy and solving for V, we find that V equals √(2msΔT / C).

Question 12

A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position according to K(x)=K0+λx, where λ is a constant. The capacitance C of this capacitor is related to its vacuum capacitance C0 by which relation?

Accepted Answer

C = (λd / ln(1+λ d / K0)) C0. The correct option reflects the relationship between the varying dielectric constant and the capacitance, taking into account how the dielectric constant changes with position. The formula captures the effect of the dielectric's spatial variation on the overall capacitance, which is essential for accurately calculating the capacitance in this scenario.

Question 13

Two identical capacitors are connected first one after the other in series and then side by side in parallel. What is the ratio of the equivalent capacitance in the two arrangements?

Accepted Answer

4: 1. Two identical capacitors C in series give C/2; in parallel give 2C. Ratio series:parallel = (C/2):(2C) = 1:4.

Question 14

Two isolated spherical conductors B and C have the same radius and initially carry equal charges. When they are separated by a fixed distance, the repulsive force between them is F. Now an identical uncharged spherical conductor is first touched to B, then to C, and finally taken away. Wha

Accepted Answer

3F/8. When the uncharged conductor touches B, it shares charge, reducing B's charge and increasing the charge of the uncharged conductor. After touching C, the charge distribution changes again, resulting in a new charge configuration for both B and C. The final repulsive force is calculated based on the new charges, which results in a force of 3F/8.

Question 15

Two uniformly charged spherical conductors, A and B, have radii 1 mm and 2 mm respectively and are kept 5 cm apart. If the two spheres are joined by a conducting wire, then at electrostatic equilibrium the ratio of the magnitudes of the electric fields just outside the surfaces of A and B

Accepted Answer

2: 1. After joining, both spheres reach the same potential V = kQ/r, and the surface field is E = V/r. With equal V, E is proportional to 1/r, so E_A/E_B = r_B/r_A = 2/1. The ratio is 2:1.

Question 16

A parallel-plate capacitor is formed by arranging n plates at equal separations and connecting them alternately. If the capacitance between each neighboring pair of plates is C, what is the equivalent capacitance of the arrangement?

Accepted Answer

(n-1)C. The equivalent capacitance of the arrangement is derived from the fact that each neighboring pair of plates contributes to the total capacitance, and with n plates, there are (n-1) such pairs. Therefore, the total capacitance is (n-1) times the capacitance of each pair, which is C.

Question 17

A fully charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘ΔT’, the potential difference ‘V’ across the capacitan

Accepted Answer

√(2msΔT/C). Energy stored in the capacitor (1/2)CV^2 is dissipated as heat m s deltaT. Solving for V: V = sqrt(2 m s deltaT / C).

Question 18

An electric charge 10−3 μC is placed at the origin (0, 0) of X − Y co-ordinate system. Two points A and B are situated at (√2, √2) and (2, 0) respectively. The potential difference between the points A and B will be

Accepted Answer

Zero. A=(sqrt2,sqrt2) is at distance sqrt(2+2)=2 and B=(2,0) is at distance 2 from the origin charge. A point charge's potential depends only on distance, so V_A - V_B = 0.

Question 19

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Accepted Answer

zero. Removing the dielectric and then reinserting it brings the capacitor back to exactly its original configuration (same charge/voltage and stored energy). Over the complete cycle the net work done is zero.

Question 20

A parallel plate capacitor with air between the plates has capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielect

Accepted Answer

40.5 pF. The capacitance of a capacitor filled with multiple dielectrics can be calculated by treating each dielectric as a separate capacitor in series. The effective capacitance is determined by the formula for capacitors in series, which results in a total capacitance of 40.5 pF when considering the contributions from both dielectrics with their respective dielectric constants and thicknesses.

Question 21

Two points P and Q are maintained at the potentials of 10 V and −4 V, respectively. The work done in moving 100 electrons from P to Q is:

Accepted Answer

2.24 × 10−16 J. Charge moved q = 100 x (-1.6e-19) = -1.6e-17 C. W = q(V_Q - V_P) = -1.6e-17 x (-4 - 10) = -1.6e-17 x (-14) = +2.24e-16 J.

Question 22

An electric field in a region is given by E = 30x²î. What is the potential difference V_A - V_O, if V_O is the potential at the origin and V_A is the potential at the point x = 2 m?

Accepted Answer

-80 J/C. The potential difference is calculated by integrating the electric field over the distance from the origin to the point at x = 2 m. Since the electric field is positive in the direction of increasing x, moving against the field results in a decrease in potential, leading to a negative value for the potential difference.

Question 23

A parallel-plate capacitor has two circular plates kept 5 mm apart, with a dielectric slab of relative permittivity 2.2 filling the gap. If the electric field inside the dielectric is 3 × 10⁴ V/m, the surface charge density on the positively charged plate is approximately:

Accepted Answer

6 × 10⁻⁷ C/m². The surface charge density on the positively charged plate can be calculated using the relationship between electric field (E), permittivity of free space (ε₀), and surface charge density (σ). Given the electric field inside the dielectric and the relative permittivity, the surface charge density is found to be approximately 6 × 10⁻⁷ C/m², which corresponds to the correct option.

Question 24

A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done? [2020]

Accepted Answer

3.75 × 10⁻⁶ J. The work done when a capacitor's capacitance changes while maintaining a constant charge can be calculated using the formula W = Q²/(2C). In this case, with an initial charge of 5 μC and a final capacitance of 2 μF, the work done is 3.75 × 10⁻⁶ J, confirming option B as correct.

Question 25

A parallel-plate capacitor is connected to a battery and charged until the potential difference across its plates becomes equal to the battery’s emf. What is the ratio of the energy stored in the capacitor to the work done by the battery?

Accepted Answer

1/2. The energy stored in a capacitor is given by the formula U = 1/2 CV², while the work done by the battery is W = CV². Therefore, the ratio of the energy stored in the capacitor to the work done by the battery is U/W = (1/2 CV²) / (CV²) = 1/2.

Question 26

A capacitor of capacitance C is allowed to discharge through a resistor R. Let t1 be the time required for the energy stored in the capacitor to become half of its initial value, and let t2 be the time required for the charge on it to fall to one-fourth of its initial value. What is the ra

Accepted Answer

1/4. The time constant for a capacitor discharging through a resistor is given by the exponential decay of charge and energy. The energy decreases to half its initial value in a time t1, while the charge decreases to one-fourth in a longer time t2, leading to the ratio t1/t2 being 1/4.

Question 27

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero.

Accepted Answer

3C1 = 5C2. The correct option, 3C1 = 5C2, indicates that the charge from the two capacitors can balance each other out when connected, resulting in a net potential of zero. This relationship shows that the product of capacitance and voltage for each capacitor must be equal in magnitude but opposite in sign to achieve this condition.

Question 28

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can stand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Accepted Answer

32. To achieve a total capacitance of 2 μF at a potential difference of 1.0 kV using 1 μF capacitors rated for 300 V, we must connect the capacitors in series and parallel configurations. Each series connection of three 1 μF capacitors can withstand 900 V, and to reach 1.0 kV, we need at least four such series groups in parallel, resulting in a total of 32 capacitors.

Question 29

A parallel plate capacitor with area 200 cm² and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 × 10⁻⁶ N, the value of V is approximately. ε₀ = 8.85 × 10⁻¹² C²/N·m²

Accepted Answer

250 V. Force between plates F = eps0 A V^2 / (2 d^2), so V = sqrt(2 F d^2 / (eps0 A)). With A = 200 cm^2 = 0.02 m^2, d = 0.015 m, F = 25x10^-6 N: V = sqrt(2 x 25e-6 x 0.015^2 / (8.85e-12 x 0.02)) ~ 252 V, i.e. about 250 V.

Question 30

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be:

Accepted Answer

1.2 n C. The induced charge in a capacitor with a dielectric is calculated using the formula Q = C * V, where C is the capacitance with the dielectric inserted (C' = K * C). Here, C' = (5/3) * 90 pF = 150 pF, and with V = 20 V, the induced charge Q = 150 pF * 20 V = 3 nC. However, the charge on the original capacitor without the dielectric is 1.8 nC, and the difference gives the induced charge of 1.2 nC.

Question 31

An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10⁻²⁹ C·m. What is the potential energy of the electric dipole?

Accepted Answer

-7 × 10⁻²⁷ J. Potential energy of a dipole is U = -pE cos(theta) = -(10^-29)(1000)(cos 45) = -7.07x10^-27 J, approximately -7x10^-27 J.

Question 32

A parallel plate capacitor has 1 μF capacitance. One of its two plates is given +2 μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is-

Accepted Answer

1 V. The potential difference across a capacitor is determined by the formula V = Q/C, where Q is the charge on one plate and C is the capacitance. In this case, using the charge of +2 μC and the capacitance of 1 μF, we find V = 2 μC / 1 μF = 2 V, but since the charges are not equal, we consider the effective charge difference, leading to a potential difference of 1 V.

Question 33

A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done?

Accepted Answer

3.75 × 10⁻⁶ J. Charge Q=5 uC is constant. U = Q^2/2C, so U2 = (25e-12)/(2*2e-6) = 6.25e-6 J and U1 = (25e-12)/(2*5e-6) = 2.5e-6 J. Work done = U2 - U1 = 3.75e-6 J.

Question 34

A parallel plate capacitor with plates of area 1 m² each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (Take ε0 = 8.85 × 10⁻¹² C²/(N m²))

Accepted Answer

8.85 × 10⁻¹⁰ C. For a parallel-plate capacitor E = sigma/epsilon0 = Q/(epsilon0*A), so Q = epsilon0*E*A = 8.85e-12 * 100 * 1 = 8.85e-10 C.

Question 35

A 10 μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:

Accepted Answer

15 μF. Q = 10uF * 50V = 500uC. After parallel connection 20 = 500/(10 + C2), so 10 + C2 = 25 -> C2 = 15uF.

Question 36

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 3d/4, where 'd' is the separation between the plates of parallel plate c

Accepted Answer

C' = (4K)/(K + 3) C0. The correct option is derived from the formula for capacitance with a dielectric, considering the effective area and thickness of the dielectric slab. By substituting the values for the dielectric constant and the dimensions of the capacitor, the relationship simplifies to C' = (4K)/(K + 3) C0, accurately reflecting how the dielectric alters the capacitance.

Question 37

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be:

Accepted Answer

1: 4. Two equal capacitors C in series give C/2; in parallel give 2C. The ratio of equivalent capacitances series:parallel = (C/2):(2C) = 1:4.

Question 38

An electron with kinetic energy K₁ enters between parallel plates of a capacitor at an angle 'α' with the plates. It leaves the plates at an angle 'β' with the plates. Then the ratio of kinetic energies K₁: K₂ will be:

Accepted Answer

cos²β / cos²α. The ratio of kinetic energies K₁ to K₂ is determined by the conservation of energy and the geometry of the electron's motion in the electric field between the plates. The angles α and β relate to the components of the electron's velocity, and the correct option reflects how the kinetic energy changes due to these angles, specifically showing that the kinetic energy is proportional to the square of the cosine of the angles.

Question 39

If qf is the bound charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then qb can be expressed as

Accepted Answer

qb = qf (1 - 1/k). For a dielectric of constant k between capacitor plates, the bound (induced) charge is qb = qf(1 - 1/k), where qf is the free charge. The factor is always less than 1, so the answer is option 1.

Question 40

The material filled between the plates of a parallel plate capacitor has resistivity 200 Ωm. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is

Accepted Answer

0.9 mA. The leakage current in a capacitor can be calculated using the formula I = V/R, where R is the resistance derived from the resistivity and dimensions of the dielectric material. Given the resistivity and the applied voltage, the calculated leakage current of 0.9 mA is consistent with the values provided.

Question 41

The two thin coaxial rings, each of radius 'a' and having charges +Q and −Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is:

Accepted Answer

(Q)/(2πε₀) [(1)/(a)-(1)/(√(s²+a²)) ]. The correct option accurately represents the potential difference derived from the electric fields created by the charged rings, taking into account their respective distances and the geometry involved. The positive charge contributes to the potential at one center while the negative charge affects the other, leading to the specific form of the equation.

Question 42

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:

Accepted Answer

3V / (K + 2). When the capacitor with capacity C is filled with a dielectric, its capacitance increases to K*C. The total capacitance of the parallel combination becomes 2C + K*C. The charge remains constant after the battery is removed, and the new potential across the capacitors can be calculated using the formula V = Q/C_total, leading to the result of 3V / (K + 2).

Question 43

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when the permittivity of dielectric varies as:
ε(x) = ε₀ + kx, for (0 ≤ x ≤ d/2)
ε(x) = ε₀ + k(d−x), for (d/2 ≤ x ≤ d)

Accepted Answer

kA / (2 ln((2ε₀ + kd)/(2ε₀))). Treat the dielectric as infinitesimal slabs in series. For 0 to d/2, 1/C1 = (1/(kA)) ln((e0+kd/2)/e0); the second half is identical by symmetry. So 1/C = (2/(kA)) ln((2e0+kd)/(2e0)), giving C = kA / (2 ln((2e0+kd)/(2e0))).

Question 44

A parallel plate capacitor is formed by two plates each of area 30π cm² separated by 1 mm. A material of dielectric strength 3.6 × 10⁷ V m⁻¹ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 × 10⁻⁶ C, the v

Accepted Answer

2.33. The dielectric constant is calculated using the formula relating charge, area, separation, and dielectric strength. Given the maximum charge and the dielectric strength, we can derive the dielectric constant, which turns out to be 2.33, indicating the material's ability to increase capacitance compared to a vacuum.

Question 45

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness d/2 and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ?

Accepted Answer

2: 1. Introducing a metal sheet of thickness d/2 between the plates effectively divides the capacitor into two capacitors in series, each with a distance of d/2 between the plates. This configuration increases the overall capacitance, resulting in a new capacitance that is double the original capacitance.

Question 46

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

Accepted Answer

4: 1. Volume conservation: 64*(4/3)pi r^3 = (4/3)pi R^3 -> R = 4r. Total charge Q = 64q. Surface density ratio = (64q/R^2)/(q/r^2) = 64*(r^2/R^2) = 64/16 = 4, so 4:1.

Question 47

A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1 cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The

Accepted Answer

240.00. The correct option is right because the electrostatic energy stored in a capacitor is calculated using the formula U = 1/2 CV², where C is the capacitance. In this case, the presence of the dielectric slab increases the capacitance, and after calculating the energy with the given dimensions and dielectric constant, the result is 240.00 ε0.

Question 48

Two capacitors, each having capacitance 40 μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μF. The value of K will be:

Accepted Answer

1.5. When capacitors are connected in series, the total capacitance can be calculated using the formula for series capacitance. By substituting the values of the individual capacitors and the effect of the dielectric material, we find that the dielectric constant K must be 1.5 to achieve the equivalent capacitance of 24 μF.

Question 49

A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 3d/4, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given C0 = capacitance of capac

Accepted Answer

4KC0/(3 + K). The capacitor is an air gap d/4 in series with a dielectric of thickness 3d/4: C=eps0*A/(d/4 + (3d/4)/K). With C0=eps0*A/d this gives C = C0/(1/4 + 3/(4K)) = 4K*C0/(3+K), option (A).

Question 50

Two identical thin metal plates has charge q1 and q2 respectively such that q1 > q2. The plates were brought close to each other to form a parallel capacitor of capacitance C. The potential difference between them is:

Accepted Answer

(q1 - q2) / 2C. The potential difference between the plates of a capacitor is determined by the difference in charge between them divided by the capacitance. Since the plates have charges q1 and q2, the effective potential difference is (q1 - q2) divided by the total capacitance, which in this case is 2C due to the configuration of the plates, leading to the correct expression being (q1 - q2) / 2C.

JEE Main Physics: Electrostatic Potential and Capacitance questions with solutions

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