Neet Pyq Electric Potential And Capacitance

Neet Pyq Electric Potential And Capacitance

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In the NEET (National Eligibility cum Entrance Test), questions related to Electric Potential and Capacitance are frequently asked. This topic is crucial in electrostatics, covering concepts such as electric potential energy, potential difference, capacitance of conductors, and energy stored in capacitors.

This topic provides a detailed explanation of Electric Potential and Capacitance, along with important previous year questions (PYQ) from NEET to help students prepare effectively.

Understanding Electric Potential

What is Electric Potential?

Electric potential at a point in an electric field is defined as the work done per unit charge in bringing a positive test charge from infinity to that point without acceleration. It is given by:

V = frac{W}{q}

where:

  • V = Electric potential (Volts)
  • W = Work done (Joules)
  • q = Charge (Coulombs)

The SI unit of electric potential is Volt (V), where 1 Volt = 1 Joule/Coulomb.

Electric Potential Due to a Point Charge

For a single point charge Q at a distance r, the electric potential is given by:

V = frac{1}{4piepsilon_0} cdot frac{Q}{r}

where ε₀ (epsilon naught) is the permittivity of free space.

Electric Potential Difference

The electric potential difference between two points in an electric field is the work done in moving a unit charge from one point to another. It is given by:

Delta V = V_B – V_A = frac{W}{q}

This concept is crucial in understanding capacitors, batteries, and electric circuits.

Understanding Capacitance

What is Capacitance?

Capacitance is the ability of a conductor to store electric charge. It is defined as the ratio of charge (Q) stored to the potential difference (V) across the conductor:

C = frac{Q}{V}

where:

  • C = Capacitance (Farad, F)
  • Q = Charge (Coulombs)
  • V = Potential difference (Volts)

Capacitance of a Parallel Plate Capacitor

For a parallel plate capacitor with plate area A and separation d, the capacitance is:

C = frac{epsilon_0 epsilon_r A}{d}

where ε₀ is the permittivity of free space and ε_r is the relative permittivity (dielectric constant).

Capacitors in Series and Parallel

1. Capacitors in Series

When capacitors are connected in series, the equivalent capacitance is given by:

frac{1}{C_{text{eq}}} = frac{1}{C_1} + frac{1}{C_2} + frac{1}{C_3} + dots

2. Capacitors in Parallel

For capacitors in parallel, the total capacitance is:

C_{text{eq}} = C_1 + C_2 + C_3 + dots

Parallel combination increases total capacitance, while series combination decreases it.

Energy Stored in a Capacitor

The energy (U) stored in a capacitor is given by:

U = frac{1}{2} C V^2

This energy is stored in the electric field between the capacitor plates and is used in various electronic circuits.

Important NEET PYQ on Electric Potential and Capacitance

1. Question on Electric Potential Difference

Q1: Two point charges +Q and −Q are placed at a distance 2a apart. The electric potential at the midpoint is:

(A) Zero
**(B) frac{Q}{4piepsilon_0 a} **
**(C) frac{2Q}{4piepsilon_0 a} **
**(D) frac{-Q}{4piepsilon_0 a} **

Solution:

The potential due to a charge at a distance r is given by:

V = frac{1}{4piepsilon_0} cdot frac{Q}{r}

Since the charges are equal and opposite, their potentials cancel at the midpoint, making the total potential zero.

Answer: (A) Zero

2. Question on Capacitance of a Parallel Plate Capacitor

Q2: The capacitance of a parallel plate capacitor increases when:

(A) Plate separation increases
(B) Dielectric constant increases
(C) Charge on plates decreases
(D) Potential difference decreases

Solution:

From the capacitance formula:

C = frac{epsilon_0 epsilon_r A}{d}
  • Increasing ε_r (dielectric constant) increases capacitance.
  • Increasing d (plate separation) decreases capacitance.

Answer: (B) Dielectric constant increases

3. Question on Capacitors in Series

Q3: Three capacitors of 4μF, 6μF, and 12μF are connected in series. Find the equivalent capacitance.

Solution:

Using the series capacitance formula:

frac{1}{C_{text{eq}}} = frac{1}{4} + frac{1}{6} + frac{1}{12}

Solving:

frac{1}{C_{text{eq}}} = frac{3}{12} + frac{2}{12} + frac{1}{12} = frac{6}{12}
C_{text{eq}} = frac{12}{6} = 2μF

Answer: 2μF

4. Question on Energy Stored in a Capacitor

Q4: A 10μF capacitor is charged to 100V. What is the energy stored?

Solution:

Using the formula:

U = frac{1}{2} C V^2
U = frac{1}{2} times 10 times 10^{-6} times (100)^2
U = frac{1}{2} times 10^{-5} times 10000
U = 0.05 J

Answer: 0.05 Joules

Tips for Solving NEET Questions on Electric Potential and Capacitance

  1. Understand Formulas – Memorize key equations for potential, capacitance, and energy storage.
  2. Apply Concepts to Problems – Use theoretical concepts to solve numerical problems efficiently.
  3. Practice PYQs – Solve previous year questions to identify common question patterns.
  4. Use Approximate Values – For quick calculations, use approximate values like ε₀ ≈ 8.85 × 10⁻¹² F/m.
  5. Check Units – Ensure proper unit conversions while solving numerical problems.

Electric Potential and Capacitance is a significant topic in NEET physics. Understanding concepts like electric potential, capacitance, energy storage, and capacitor combinations is essential for scoring well. Regular practice of previous year questions (PYQs) helps improve problem-solving speed and accuracy.

By mastering this topic, students can confidently attempt electrostatics-based questions in NEET and achieve a high score.