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JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance
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Page 1 JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance Q1: A parallel-plate capacitor of capacitance ???? ?? ?? is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant ?? = ?? . Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 8 mC and 2.0 J B. 4 mC and 0.2 J C. 2 mC and 0.2 J D. 2 mC and 0.4 J Ans: B Solution: Given, ?? = 2, ?? = 40???? , ?? = 100 ?? ??? = ( ???? ) ?? - ???? (As ?? = ???? ) = ( ?? - 1) ???? = ( 2 - 1)× 40 × 10 -6 × 100 = 4 × 10 -3 ?? ??? = 4???? ??? = 1 2 ?? ' ?? 2 - 1 2 ?? ?? 2 = 1 2 ?? 2 ( ?? ' - ?? ) = 1 2 ?? 2 ( ???? - ?? ) = 1 2 ?? 2 ?? ( ?? - 1) = 1 2 × 10 4 × 40 × 10 -6 ( 2 - 1) = ??? = 2 × 10 -1 = 0.2 J ? ??? = 0.2 J Q2: An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity ???? ?? ?? /?? . If the magnitude of the electric field between the plates is ?? . ?? ?? /???? , then the vertical component of velocity of electron is(mass of electron = ?? . ?? × ???? -???? ???? and charge of electron = ?? . ?? × ???? -???? ?? ) Page 2 JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance Q1: A parallel-plate capacitor of capacitance ???? ?? ?? is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant ?? = ?? . Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 8 mC and 2.0 J B. 4 mC and 0.2 J C. 2 mC and 0.2 J D. 2 mC and 0.4 J Ans: B Solution: Given, ?? = 2, ?? = 40???? , ?? = 100 ?? ??? = ( ???? ) ?? - ???? (As ?? = ???? ) = ( ?? - 1) ???? = ( 2 - 1)× 40 × 10 -6 × 100 = 4 × 10 -3 ?? ??? = 4???? ??? = 1 2 ?? ' ?? 2 - 1 2 ?? ?? 2 = 1 2 ?? 2 ( ?? ' - ?? ) = 1 2 ?? 2 ( ???? - ?? ) = 1 2 ?? 2 ?? ( ?? - 1) = 1 2 × 10 4 × 40 × 10 -6 ( 2 - 1) = ??? = 2 × 10 -1 = 0.2 J ? ??? = 0.2 J Q2: An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity ???? ?? ?? /?? . If the magnitude of the electric field between the plates is ?? . ?? ?? /???? , then the vertical component of velocity of electron is(mass of electron = ?? . ?? × ???? -???? ???? and charge of electron = ?? . ?? × ???? -???? ?? ) JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 1 × 10 6 m/s B. 16 × 10 6 m/s C. 16 × 10 4 m/s D. 0 Ans: B Solution: No force in horizontal direction, i.e., ?? ?? = 0 ? ?? ?? = 0 So, ?? ?? = constant hence, ?? = ?? ?? ?? …. (1) Now, for y -direction, using Ist equation of motion, ?? ?? = ?? ?? + ?? ?? ?? ? ?? ?? = 0 + ???? ?? ( ?? ?? ?? ) (from (1) and usiung ?? = ???? and ?? = ?? ?? ) ? ?? ?? = 1.6 × 10 -19 × 9.1 × 10 2 × 0.1 9.1 × 10 -31 × 10 6 = 1.6 × 10 7 m/s ? ?? ?? = 16 × 10 6 m/s. Q3: Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and ?? stand for unit of mass, length, time and charge, JEE Main 2025 (Online) 22nd January Evening Shift Options: A. [F] = [CM -1 L -2 T 2 ] B. [?? ] = [?? 2 ?? -2 ?? 2 ?? 2 ] C. [F] = [C 2 M -1 L -2 T 2 ] D. [F] = [CM -2 L -2 T -2 ] Ans: C Solution: Page 3 JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance Q1: A parallel-plate capacitor of capacitance ???? ?? ?? is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant ?? = ?? . Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 8 mC and 2.0 J B. 4 mC and 0.2 J C. 2 mC and 0.2 J D. 2 mC and 0.4 J Ans: B Solution: Given, ?? = 2, ?? = 40???? , ?? = 100 ?? ??? = ( ???? ) ?? - ???? (As ?? = ???? ) = ( ?? - 1) ???? = ( 2 - 1)× 40 × 10 -6 × 100 = 4 × 10 -3 ?? ??? = 4???? ??? = 1 2 ?? ' ?? 2 - 1 2 ?? ?? 2 = 1 2 ?? 2 ( ?? ' - ?? ) = 1 2 ?? 2 ( ???? - ?? ) = 1 2 ?? 2 ?? ( ?? - 1) = 1 2 × 10 4 × 40 × 10 -6 ( 2 - 1) = ??? = 2 × 10 -1 = 0.2 J ? ??? = 0.2 J Q2: An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity ???? ?? ?? /?? . If the magnitude of the electric field between the plates is ?? . ?? ?? /???? , then the vertical component of velocity of electron is(mass of electron = ?? . ?? × ???? -???? ???? and charge of electron = ?? . ?? × ???? -???? ?? ) JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 1 × 10 6 m/s B. 16 × 10 6 m/s C. 16 × 10 4 m/s D. 0 Ans: B Solution: No force in horizontal direction, i.e., ?? ?? = 0 ? ?? ?? = 0 So, ?? ?? = constant hence, ?? = ?? ?? ?? …. (1) Now, for y -direction, using Ist equation of motion, ?? ?? = ?? ?? + ?? ?? ?? ? ?? ?? = 0 + ???? ?? ( ?? ?? ?? ) (from (1) and usiung ?? = ???? and ?? = ?? ?? ) ? ?? ?? = 1.6 × 10 -19 × 9.1 × 10 2 × 0.1 9.1 × 10 -31 × 10 6 = 1.6 × 10 7 m/s ? ?? ?? = 16 × 10 6 m/s. Q3: Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and ?? stand for unit of mass, length, time and charge, JEE Main 2025 (Online) 22nd January Evening Shift Options: A. [F] = [CM -1 L -2 T 2 ] B. [?? ] = [?? 2 ?? -2 ?? 2 ?? 2 ] C. [F] = [C 2 M -1 L -2 T 2 ] D. [F] = [CM -2 L -2 T -2 ] Ans: C Solution: The capacitance (in farads) is defined as the ratio of charge to potential difference. Let's go through the steps: Capacitance is given by: ?? = ?? ?? where: ?? is the charge with dimensional symbol ?? . ?? is the potential difference. Voltage (potential difference) is defined as energy per unit charge: ?? = ?? ?? where: ?? is energy with dimensions: [?? ] = [?? ?? 2 ?? -2 ] Therefore, the dimensions of voltage are: [?? ] = [?? ?? 2 ?? -2 ] [?? ] Now, substitute this back into the expression for capacitance: [?? ] = [?? ] [?? ?? 2 ?? -2 ]/[?? ] = [?? ] 2 ?? ?? 2 ?? -2 Simplifying gives: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Comparing with the given options, we see that Option C is: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Thus, the correct dimensional formula for the capacitance in farads is given by Option C. Q4: Identify the valid statements relevant to the given circuit at the instant when the key is closed. Page 4 JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance Q1: A parallel-plate capacitor of capacitance ???? ?? ?? is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant ?? = ?? . Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 8 mC and 2.0 J B. 4 mC and 0.2 J C. 2 mC and 0.2 J D. 2 mC and 0.4 J Ans: B Solution: Given, ?? = 2, ?? = 40???? , ?? = 100 ?? ??? = ( ???? ) ?? - ???? (As ?? = ???? ) = ( ?? - 1) ???? = ( 2 - 1)× 40 × 10 -6 × 100 = 4 × 10 -3 ?? ??? = 4???? ??? = 1 2 ?? ' ?? 2 - 1 2 ?? ?? 2 = 1 2 ?? 2 ( ?? ' - ?? ) = 1 2 ?? 2 ( ???? - ?? ) = 1 2 ?? 2 ?? ( ?? - 1) = 1 2 × 10 4 × 40 × 10 -6 ( 2 - 1) = ??? = 2 × 10 -1 = 0.2 J ? ??? = 0.2 J Q2: An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity ???? ?? ?? /?? . If the magnitude of the electric field between the plates is ?? . ?? ?? /???? , then the vertical component of velocity of electron is(mass of electron = ?? . ?? × ???? -???? ???? and charge of electron = ?? . ?? × ???? -???? ?? ) JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 1 × 10 6 m/s B. 16 × 10 6 m/s C. 16 × 10 4 m/s D. 0 Ans: B Solution: No force in horizontal direction, i.e., ?? ?? = 0 ? ?? ?? = 0 So, ?? ?? = constant hence, ?? = ?? ?? ?? …. (1) Now, for y -direction, using Ist equation of motion, ?? ?? = ?? ?? + ?? ?? ?? ? ?? ?? = 0 + ???? ?? ( ?? ?? ?? ) (from (1) and usiung ?? = ???? and ?? = ?? ?? ) ? ?? ?? = 1.6 × 10 -19 × 9.1 × 10 2 × 0.1 9.1 × 10 -31 × 10 6 = 1.6 × 10 7 m/s ? ?? ?? = 16 × 10 6 m/s. Q3: Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and ?? stand for unit of mass, length, time and charge, JEE Main 2025 (Online) 22nd January Evening Shift Options: A. [F] = [CM -1 L -2 T 2 ] B. [?? ] = [?? 2 ?? -2 ?? 2 ?? 2 ] C. [F] = [C 2 M -1 L -2 T 2 ] D. [F] = [CM -2 L -2 T -2 ] Ans: C Solution: The capacitance (in farads) is defined as the ratio of charge to potential difference. Let's go through the steps: Capacitance is given by: ?? = ?? ?? where: ?? is the charge with dimensional symbol ?? . ?? is the potential difference. Voltage (potential difference) is defined as energy per unit charge: ?? = ?? ?? where: ?? is energy with dimensions: [?? ] = [?? ?? 2 ?? -2 ] Therefore, the dimensions of voltage are: [?? ] = [?? ?? 2 ?? -2 ] [?? ] Now, substitute this back into the expression for capacitance: [?? ] = [?? ] [?? ?? 2 ?? -2 ]/[?? ] = [?? ] 2 ?? ?? 2 ?? -2 Simplifying gives: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Comparing with the given options, we see that Option C is: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Thus, the correct dimensional formula for the capacitance in farads is given by Option C. Q4: Identify the valid statements relevant to the given circuit at the instant when the key is closed. A. There will be no current through resistor ?? . B. There will be maximum current in the connecting wires. C. Potential difference between the capacitor plates A and B is minimum. D. Charge on the capacitor plates is minimum. Choose the correct answer from the options given below: JEE Main 2025 (Online) 23rd January Morning Shift Options: A. A, C Only B. C, D Only C. A, B, D Only D. B, C, D Only Ans: D Solution: Initially, when the key in the circuit is closed, the capacitor behaves like a short circuit. This means: Maximum Current: The current flowing through the circuit will be at its maximum since there is effectively no resistance offered by the capacitor, as it behaves like a wire. Current Through R: Since the current is flowing through the circuit, including any resistors in series, the statement that there is no current through resistor ?? is incorrect. Charge on the Capacitor: The charge stored on the capacitor plates is initially zero because it behaves as a short circuit. Potential Difference: The potential difference across the capacitor is also zero at this instant because it is fully discharged. Therefore, the valid statements at this instant are that the current in the circuit is at a maximum, the charge on the capacitor is zero, and the potential difference across the capacitor is zero. Q5: A parallel plate capacitor was made with two rectangular plates, each with a length of ?? = ?? ???? and breath of ?? = ?? ???? . The distance between the plates is ?? ?? ?? . Out of the following, which are the ways to increase the capacitance by a factor of 10 ? Page 5 JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance Q1: A parallel-plate capacitor of capacitance ???? ?? ?? is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant ?? = ?? . Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 8 mC and 2.0 J B. 4 mC and 0.2 J C. 2 mC and 0.2 J D. 2 mC and 0.4 J Ans: B Solution: Given, ?? = 2, ?? = 40???? , ?? = 100 ?? ??? = ( ???? ) ?? - ???? (As ?? = ???? ) = ( ?? - 1) ???? = ( 2 - 1)× 40 × 10 -6 × 100 = 4 × 10 -3 ?? ??? = 4???? ??? = 1 2 ?? ' ?? 2 - 1 2 ?? ?? 2 = 1 2 ?? 2 ( ?? ' - ?? ) = 1 2 ?? 2 ( ???? - ?? ) = 1 2 ?? 2 ?? ( ?? - 1) = 1 2 × 10 4 × 40 × 10 -6 ( 2 - 1) = ??? = 2 × 10 -1 = 0.2 J ? ??? = 0.2 J Q2: An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity ???? ?? ?? /?? . If the magnitude of the electric field between the plates is ?? . ?? ?? /???? , then the vertical component of velocity of electron is(mass of electron = ?? . ?? × ???? -???? ???? and charge of electron = ?? . ?? × ???? -???? ?? ) JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 1 × 10 6 m/s B. 16 × 10 6 m/s C. 16 × 10 4 m/s D. 0 Ans: B Solution: No force in horizontal direction, i.e., ?? ?? = 0 ? ?? ?? = 0 So, ?? ?? = constant hence, ?? = ?? ?? ?? …. (1) Now, for y -direction, using Ist equation of motion, ?? ?? = ?? ?? + ?? ?? ?? ? ?? ?? = 0 + ???? ?? ( ?? ?? ?? ) (from (1) and usiung ?? = ???? and ?? = ?? ?? ) ? ?? ?? = 1.6 × 10 -19 × 9.1 × 10 2 × 0.1 9.1 × 10 -31 × 10 6 = 1.6 × 10 7 m/s ? ?? ?? = 16 × 10 6 m/s. Q3: Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and ?? stand for unit of mass, length, time and charge, JEE Main 2025 (Online) 22nd January Evening Shift Options: A. [F] = [CM -1 L -2 T 2 ] B. [?? ] = [?? 2 ?? -2 ?? 2 ?? 2 ] C. [F] = [C 2 M -1 L -2 T 2 ] D. [F] = [CM -2 L -2 T -2 ] Ans: C Solution: The capacitance (in farads) is defined as the ratio of charge to potential difference. Let's go through the steps: Capacitance is given by: ?? = ?? ?? where: ?? is the charge with dimensional symbol ?? . ?? is the potential difference. Voltage (potential difference) is defined as energy per unit charge: ?? = ?? ?? where: ?? is energy with dimensions: [?? ] = [?? ?? 2 ?? -2 ] Therefore, the dimensions of voltage are: [?? ] = [?? ?? 2 ?? -2 ] [?? ] Now, substitute this back into the expression for capacitance: [?? ] = [?? ] [?? ?? 2 ?? -2 ]/[?? ] = [?? ] 2 ?? ?? 2 ?? -2 Simplifying gives: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Comparing with the given options, we see that Option C is: [?? ] = [?? 2 ?? -1 ?? -2 ?? 2 ] Thus, the correct dimensional formula for the capacitance in farads is given by Option C. Q4: Identify the valid statements relevant to the given circuit at the instant when the key is closed. A. There will be no current through resistor ?? . B. There will be maximum current in the connecting wires. C. Potential difference between the capacitor plates A and B is minimum. D. Charge on the capacitor plates is minimum. Choose the correct answer from the options given below: JEE Main 2025 (Online) 23rd January Morning Shift Options: A. A, C Only B. C, D Only C. A, B, D Only D. B, C, D Only Ans: D Solution: Initially, when the key in the circuit is closed, the capacitor behaves like a short circuit. This means: Maximum Current: The current flowing through the circuit will be at its maximum since there is effectively no resistance offered by the capacitor, as it behaves like a wire. Current Through R: Since the current is flowing through the circuit, including any resistors in series, the statement that there is no current through resistor ?? is incorrect. Charge on the Capacitor: The charge stored on the capacitor plates is initially zero because it behaves as a short circuit. Potential Difference: The potential difference across the capacitor is also zero at this instant because it is fully discharged. Therefore, the valid statements at this instant are that the current in the circuit is at a maximum, the charge on the capacitor is zero, and the potential difference across the capacitor is zero. Q5: A parallel plate capacitor was made with two rectangular plates, each with a length of ?? = ?? ???? and breath of ?? = ?? ???? . The distance between the plates is ?? ?? ?? . Out of the following, which are the ways to increase the capacitance by a factor of 10 ? A. ?? = ???? ???? , ?? = ?? ???? , ?? = ?? ?? ?? B. ?? = ?? ???? , ?? = ?? ???? , ?? = ???? ?? ?? C. ?? = ?? ???? , ?? = ?? ???? , ?? = ?? ?? ?? D. ?? = ?? ???? , ?? = ?? ???? , ?? = ???? ?? ?? E. ?? = ?? ???? , ?? = ?? ???? , ?? = ?? ?? ?? Choose the correct answer from the options given below: JEE Main 2025 (Online) 24th January Morning Shift Options: A. C only B. A only C. B and D only D. C and E only Ans: D Solution: The capacitance of a parallel plate capacitor is given by ?? = ?? 0 ?? ?? , where: ?? 0 is the permittivity of free space, ?? is the area of a plate, and ?? is the separation between the plates. A change in the capacitor's dimensions will change its capacitance by a factor of ?? new ?? initial = ?? new ?? initial · ?? initial ?? new . The initial dimensions are: Length: ?? = 3 cm = 0.03 m, Breadth: ?? = 1 cm = 0.01 m, Hence, the original area is ?? initial = 0.03 × 0.01 = 3 × 10 -4 m 2 , Plate separation: ?? initial = 3?? m = 3 × 10 -6 m. For each modification, we compare the new factor: Case C: New dimensions: ?? = 6 cm = 0.06 m, ?? = 5 cm = 0.05 m. New area: ?? new = 0.06 × 0.05 = 3 × 10 -3 m 2 . Ratio of areas: ?? new ?? initial = 3×10 -3 3×10 -4 = 10. The plate separation is unchanged: ?? initial ?? new = 3×10 -6 3×10 -6 = 1. Overall factor: 10 × 1 = 10.Read More
FAQs on JEE Main Previous Year Questions (2025): Electrostatic Potential and Capacitance
| 1. What is electrostatic potential and how is it calculated? |  |
Ans. Electrostatic potential, often referred to as electric potential, is the amount of work done in bringing a unit positive charge from infinity to a point in an electric field without any acceleration. It is calculated using the formula V = k * Q / r, where V is the electrostatic potential, k is Coulomb's constant (approximately 8.99 × 10⁹ N·m²/C²), Q is the charge creating the field, and r is the distance from the charge to the point of interest.
| 2. What is capacitance and how is it defined? | |
Ans. Capacitance is the ability of a system to store an electric charge. It is defined as the charge stored per unit potential difference across the plates of a capacitor. The formula for capacitance (C) is C = Q / V, where Q is the charge stored and V is the voltage across the capacitor. The unit of capacitance is the farad (F).
| 3. How do parallel plate capacitors work and what factors affect their capacitance? | |
Ans. Parallel plate capacitors consist of two conductive plates separated by a dielectric material. The capacitance of a parallel plate capacitor is given by the formula C = ε₀ * A / d, where ε₀ is the permittivity of free space, A is the area of one of the plates, and d is the separation distance between the plates. The capacitance increases with a larger plate area and a smaller separation distance, as well as with the use of a dielectric material with a higher permittivity.
| 4. What is the difference between series and parallel combinations of capacitors? | |
Ans. In a series combination of capacitors, the total capacitance (Cₜ) is given by the reciprocal of the sum of the reciprocals of individual capacitances: 1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃. This results in a lower total capacitance. In contrast, for capacitors in parallel, the total capacitance is simply the sum of individual capacitances: Cₜ = C₁ + C₂ + C₃. This configuration yields a higher total capacitance than any of the individual capacitors.
| 5. How does a dielectric material affect the capacitance of a capacitor? | |
Ans. A dielectric material, when placed between the plates of a capacitor, increases the capacitance by reducing the electric field between the plates for a given charge. This reduction allows the capacitor to store more charge for the same applied voltage. The capacitance with a dielectric is given by C' = k * C, where C' is the new capacitance, k is the dielectric constant of the material, and C is the original capacitance without the dielectric. The higher the dielectric constant, the greater the increase in capacitance.
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