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Electrostatic Potential and Capacitance appeared 3 times across 1 year — 6.7% of Physics. This question is from Combination of Capacitors.

Year 2024 Total
Questions 3 3
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Five capacitors of capacitances C₁ = C₂ = C₃ = C₄ = 10 ~μ F and C₅ = 2.5 ~μ F are connected as shown, along with a battery of 50 ~V.
Circuit diagram with five capacitors and a 50V battery for Q15
Circuit containing series and parallel combination of C1, C2, C3, C4 and C5 with 50V battery.
The equivalent capacitance and the charges on each capacitor respectively are:

Solution & Explanation

Related Formula
Cₛₑᵣᵢₑₛ = (1)/((1)/(C₁) + (1)/(C₂) + ⋯) Cparallel = C₁ + C₂

Q = CV

Core Logic

The schematic

Simplified parallel block diagram for Q15
Circuit containing series and parallel combination of C1, C2, C3, C4 and C5 with 50V battery.
shows C₁, C₂, C₃, C₄ are in series with each other. This entire series branch is in parallel with C₅.

Equivalent capacitance of the series branch (C₁ to C₄):

(1)/(CS) = (1)/(10) + (1)/(10) + (1)/(10) + (1)/(10) = (4)/(10) CS = 2.5 ~μ F

Total equivalent capacitance:

Ceq = CS + C₅ = 2.5 + 2.5 = 5 ~μ F
Step 1: Find Charges

The voltage across the series branch is 50 ~V. Charge on each capacitor in the series branch (C₁, C₂, C₃, C₄):

q₁ = q₂ = q₃ = q₄ = CS × V = 2.5 × 50 = 125 ~μ C

The voltage across C₅ is also 50 ~V. Charge on C₅:

q₅ = C₅ × V = 2.5 × 50 = 125 ~μ C
Pattern Recognition

Since CS and C₅ have identical capacitance (2.5 ~μ F) and are in parallel, they draw identical charge. Thus, every capacitor in the array ends up with 125 ~μ C.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Reference Study Guides

Electrostatic Potential And Capacitance

More Electrostatic Potential and Capacitance Previous-Year Questions

Q31 neet_2026_03_may_morning Energy Stored in a Capacitor
Consider two uncharged capacitors of equal capacitance 200 ~pF. One of them is charged by a 100 ~V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:
  • A. 1.0 × 10⁻⁶ ~J
  • B. 0.5 × 10⁻⁶ ~J
  • C. 0.5 ~J
  • D. 1.0 ~J

Solution

Related Formula
Δ U = (1)/(2) (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²
Core Logic

When a charged capacitor is connected to an uncharged capacitor, charge redistributes until both reach a common potential, resulting in a loss of electrostatic energy (dissipated as heat/electromagnetic radiation). Given: C₁ = 200 ~pF = 200 × 10⁻¹² ~F C₂ = 200 ~pF = 200 × 10⁻¹² ~F V₁ = 100 ~V V₂ = 0 ~V (since it is uncharged)

Step 1: Calculate Energy Loss
Energy loss = (1)/(2) ( (200 × 200)/(200 + 200) ) × 10⁻¹² × (100 - 0)² Energy loss = (1)/(2) ( (40000)/(400) ) × 10⁻¹² × 10⁴ Energy loss = (1)/(2) × 100 × 10⁻¹² × 10⁴ Energy loss = (1)/(2) × 10⁶ × 10⁻¹² = 0.5 × 10⁻⁶ ~J
Pattern Recognition

When a capacitor C charged to V is connected across an identical uncharged capacitor C, exactly half of the initial stored energy is lost in redistribution.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q41 neet_2026_03_may_morning Electrostatics of Conductors
Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor. Choose the correct answer from the options given below:
  • A. C, D and E only
  • B. A, B and D only
  • C. A, C and D only
  • D. A, C and E only

Solution

Core Logic

Evaluate each statement against the rules of electrostatics for conductors:

A. True. In electrostatic equilibrium, the electric field inside a solid or hollow conductor is precisely zero.

B. False. The electric field immediately outside the surface of a conductor is E = (σ)/(ε₀). It directly depends on the local surface charge density (σ).

C. True. Due to Gauss's Law (E = 0 inside), any excess charge must reside exclusively on the outer surface.

D. True. If the field had a tangential component, charges would flow along the surface, violating the static assumption. Thus, E must be strictly perpendicular (normal).

E. False. Since E = 0 inside, the potential (V) must be constant, but not necessarily zero. It is equal to the potential at the surface.

Step 1: Final Conclusion

Only statements A, C, and D are correct.

Pattern Recognition

Watch for the "potential is zero" trap (Statement E). Potential is constant inside a conductor, but it only equals zero if the conductor is grounded.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

More Electrostatic Potential and Capacitance Questions — neet_2026_03_may_morning

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