JEE Main Capacitor Previous Year Questions with Solutions

Capacitor is a device which is used to store energy. The capacitor is an arrangement in which an insulator separates two conductors kept near each other. The insulating medium of the capacitor can be air, paper, glass or a dielectric material. Capacitors come in different shapes and sizes. The amount of electric charge that can be stored in the capacitor per unit voltage across its plates is called capacitance.

Download Capacitor Previous Year Solved Questions PDF

Application of Capacitors

Capacitors have a wide range of applications. Some applications are given below:

• Capacitors are used in computers to store a large amount of data.
• Sensitive components of electric surges are protected by using capacitors as filters.
• Air humidity, fuel levels, etc can be measured by using capacitors as sensors.
• Capacitor coupling is the process of effectively blocking DC current and passing AC current using capacitors.
• Unwanted signals are filtered using capacitors.

JEE Main Previous Year Solved Questions on Capacitor

Q1: 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^–1, the magnitude of charge on each plate is

(Taken ε₀ = 8.85 x 10^-12 C²/N m²)

(a) 8.85 × 10^–10 C

(b) 7.85 × 10^–10 C

(c) 9.85 × 10^–10 C

(d) 6.85 × 10^–10 C

Solution

The electric field between two plates is

E = σ/ε₀ = q/Aε₀ ⇒q = EAε₀

q = (100)(1)(8.85 x 10^-12)= 8.85 × 10^–10 C

Answer: (a) 8.85 × 10^–10 C

Q2: The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is

(a) V

(b) (n+1)V/(K+n)

(c) (nV)/(K+n)

(d) V/(K+n)

Solution

For parallel combination, C_eq = C + nC = C(n + 1)

Charge on capacitor, q = C_eqV = CV(n + 1)

Now, after removing the battery, dielectric material is placed.

Then, C_eq = KC + nC = C(K + n)

New potential difference, V’ = q/C_eq =CV(n+1)/C(K+n) = (n+1)V/(K+n)

Answer: (b) (n+1)V/(K+n)

Q3: A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is

(a) C

(b) nC

(c) (n – 1)C

(d) (n + 1)C

Solution

n plates connected alternately give rise to (n – 1) capacitors connected in parallel

Resultant capacitance = (n – 1)C.

Answer: (c) (n – 1)C

Q4: Capacitance (in F) of a spherical conductor with radius 1 m is

(a) 1.1 × 10^–10

(b) 10^–6

(c) 9 × 10^–9

(d) 10^–3

Solution

C = 4πε₀R = 1/9 x 10⁹ = 1.1 x 10^-10 F

Answer: (a) 1.1 × 10^–10

Q5: If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to

(a) CV

(b) (½)nCV²

(c) CV²

(d) (1/2n)CV²

Solution

Total capacity =nC

∴ Energy =½ nCV²

Answer: (b) (½)nCV²

Q6: A parallel plate capacitor with air between the plates has a 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 dielectric constant k₂ = 6 and thickness 2d/3. The capacitance of the capacitor is now

(a) 20.25 pF

(b) 1.8 pF

(c) 45 pF

(d) 40.5 pF

Solution

C = ε₀A/d = 9 x 10^-12 F

With dielectric,C = ε₀Ak/d

C₁ = ε₀A3/(d/3) = 9C

C₂ = ε₀A6/(2d/3) = 9C

C_total = C₁C₂/(C₁ + C₂) as they are in series

C_total = (9C x 9C)/18C = (9/2)C =(9/2)(9 x 10^-12 F)

C_total = 40.5 pF

Answer: (d) 40.5 pF

Q7: 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

(a) zero

(b) 1/2 (K – 1) CV²

(c) CV²(K-1)/K

(d) (K – 1) CV²

Solution

The potential energy of a charged capacitor

U_i = q²/2C

where U_i is the initial potential energy.

If a dielectric slab is slowly introduced, the energy = q²/2KC

Once it is taken out, again the energy increases to the old value. Therefore after it is taken out, the potential energy comes back to the old value.

Total work done = zero.

Answer: (a) zero

Q8: A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

(a) 1/2

(b) 1

(c) 2

(d) 1/4

Solution

Let E be emf of the battery

Work done by the battery W = CE²

Energy stored in the capacitor

U = ½CE²

U/W = ½CE² /CE² =½

Answer: (a) 1/2

Q9: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

(a) decreases

(b) remains unchanged

(c) becomes infinite

(d) increases

Solution

With air as dielectric, C = ε₀A/d

With space partially filled, C’ = ε₀A/(d-t) = ε₀A/d = C

Aluminium is a good conductor. Its sheet introduced between the plates of a capacitor is of negligible thickness. Therefore,  capacity remains unchanged.

Answer: (b) remains unchanged

Q10: If the electric flux entering and leaving an enclosed surface respectively is Φ₁ and Φ₂, the electric charge inside the surface will be

(a) (Φ₂ – Φ₁)ε₀

(b) (Φ₁ + Φ₂)/ε₀

(c) (Φ₂ – Φ₁)/ε₀

(d) (Φ₁ + Φ₂)ε₀

Solution

According to Gauss theorem,

(Φ₂ – Φ₁) = Q/ε₀

⇒ Q = (Φ₂ – Φ₁)ε₀

The flux enters the enclosure if one has a negative charge (–q₂) and flux goes out if one has a +ve charge (+q₁). As one does not know whether Φ₁ > Φ₂, Φ₂ > Φ₁, Q = q1 ~ q2

Answer: (a) (Φ₂ – Φ₁)ε₀

Q11: A parallel plate capacitor is of area 6 cm² and a separation of 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K₁ = 10, K₂ = 12 and K₃ = 14. The dielectric constant of a material which when fully inserted in the given capacitor gives the same capacitance would be

(a) 36

(b) 12

(c) 4

(d) 14

Solution

The given system can be considered to be a parallel combination of three capacitors C₁, C₂ and C₃. The equivalent capacitance, C = C₁ + C₂ + C₃

Hence, C₁ = K₁ε₀A/3d

C₂ = K₂ε₀A/3d

C₃ = K₃ε₀A/3d

(K’ε₀A/d) = (K₁ + K₂ + K₃/3)(ε₀A/d)

Hence K’= (K₁ + K₂ + K₃/3) = (10 + 12 + 14/3) =12

Answer: (b) 12

Q12: In the figure shown below, the charge on the left plate of the 10 F capacitor is 30 μC. The charge on the right plate of the 6 F capacitor is

(a) –18 μC

(b) –12 μC

(c) +12 μC

(d) +18 μC

Solution

Let q₁ and q₂ be the charge on 6μF and 4μF respectively

q₁ + q₂ = q——-(1)

Also, (q₁/C₁) = (q₂/C₂)——-(2)

(∵ C₁ and C₂ are in parallel combination)

⇒ q₂ = (C₂/C₁) q₁= (4/6)q₁

Using (1) and (2)

(10/6)q₁ = q ⇒ q = (5/3)

⇒q₁=  (⅗)(30) = 18 μC

Answer: (d) +18 μC

Q13: Seven capacitors, each of capacitance 2 F, are to be connected in a configuration to obtain an effective capacitance of (6/13) F. Which of the combinations, shown in figures below, will achieve the desired value?

Solution

Equivalent capacitance for parallel capacitors is given by the formula

C_eq= C₁ + C₂ +C₃ ….

Therefore, 1/C_eq = 1/(C₁ + C₂ + C₃)

Equivalent capacitance for series capacitors is given by the formula

1/C_eq = 1/C₁ + 1/C₂ +1/C₃

In option “a” there are 3 parallel capacitors and 4 series capacitors

Therefore, Equivalent capacitance = 1/(2+2 +2) + (1/2) + (1/2) +(1/2) +(1/2)

= 1/6 + (1/2) + (1/2) +(1/2) +(1/2)

= 1+3+3+3+3/6

1/C_eq= 13/6 F

C_eq = (6/13) F

In option “b” there are 5 parallel capacitors and 2 series capacitors

Therefore, Equivalent capacitance = 1/(2+2 +2+2+2) + (1/2) + (1/2)

= 1/10+ (1/2) + (1/2)

= 1+5+5/10

1/C_eq= 11/10 F

C_eq = 10/11 F

In option “c” there are 2 parallel capacitors and 5 series capacitors

Therefore, Equivalent capacitance = 1/(2+2)  + (1/2) + (1/2)+ (1/2) + (1/2) +(1/2)

= 1/4+ (1/2) + (1/2)+ (1/2) + (1/2) +(1/2)

= 1+2+2+2+2+2/4

1/C_eq= 11/4 F

C_eq = (4/11) F

In option “d” there are 4 parallel capacitors and 3 series capacitors

Therefore, Equivalent capacitance = 1/(2+2+2+2)  + (1/2) + (1/2)+ (1/2)

= 1/8+ (1/2) + (1/2)+ (1/2)

= 1+4+4+4/4

1/C_eq = 13/8 F

C_{eq = (8/13) F}

Answer (a)

Q14: Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10⁶ V m^–1. The plate area is 10^–4 m². What is the dielectric constant if the capacitance is 15 pF? (given ε₀ = 8.86 × 10^–12 C² N^–1 m^–2)

(a) 3.8

(b) 8.5

(c) 6.2

(d) 4.5

Solution

C = Kε₀A/d or K = CV/ε₀AE_max

K = (15 x 10^-12 x 500)/(8.86 x 10^-12 x 10^-4 x 10⁶) = 8.5

Answer: (b) 8.5

Q15:In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is

(take 1/4πε₀ = 9 x 10⁹ Nm²C^-2)

(a) 3.0 × 10⁴ m/s

(b) 1.0 m/s

(c) 1.5 × 10² m/s

(d) 2.0 × 10³ m/s

Solution

Using work energy theorem

W_E = U_i – U_f= (½)mv² or kq₁q₂[(1/r₁) – (1/r₂)] = (½)mv²

W_E = (9 x 10⁹) x (1 x 10^-6)² {(1/10^-3) – (1/(9 x 10^-3)} = (½)mv²

⇒ (9 x 10⁹ x 2 x 10^-12)/(4 x 10^-6) = v² ⇒ v = 2.0 x 10³ ms^-1

Answer: (d) 2.0 × 10³ m/s

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