BrainIt

Physics

Electrostatic Potential and Capacitance

A Physics chapter of the NEET syllabus, and how to practise it for the computer-based exam.

Exam rules change between cycles. Confirm dates, syllabus and pattern against NTA's official site. We are not NTA.

Weightage: what the paper actually did

Electrostatic Potential and Capacitance carried 6 questions in the NEET 2026 paper. One paper is a data point, not a destiny: NTA moves weight between chapters every cycle, which is why the weightage tables deserve care. The stable facts are the paper's shape: 180 questions, 720 marks, +4 right, −1 wrong.

Sample PYQs from this chapter

Real questions NTA asked from Electrostatic Potential and Capacitance, with the answer and the working. Try before you peek.

Question 1 · NEET 2020

The capacitance of a parallel plate capacitor with air as medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
  1. A.1.77 * 10−^{-}1^{1}2^{2} C2^{2} N−^{-}1^{1} m−^{-}2^{2}
  2. B.0.44 * 10−^{-}1^{1}0^{0} C2^{2} N−^{-}1^{1} m−^{-}2^{2}
  3. C.5.00 C2^{2} N−^{-}1^{1} m−^{-}2^{2}
  4. D.0.44 * 10−^{-}1^{1}3^{3} C2^{2} N−^{-}1^{1} m−^{-}2^{2}
Answer and explanation

Answer: B

The dielectric constant is 30/6 = 5. Therefore ε = 5ε0_{0} = 4.425 * 10−^{-}1^{1}1^{1} = 0.4425 * 10−^{-}1^{1}0^{0} C2^{2} N−^{-}1^{1} m−^{-}2^{2}, matching option B. Chapter: Electrostatic Potential and Capacitance.

Question 2 · NEET 2020

A short electric dipole has a dipole moment of 16 × 10⁻⁹ C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is (1/(4πε₀) = 9 × 10⁹ N m² C⁻²)
  1. A.200 V
  2. B.400 V
  3. C.zero
  4. D.50 V
Answer and explanation

Answer: A

The electric potential due to a dipole is given by V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}. Substituting the values, V=9∗109∗16∗10−9∗cos⁡60∘(0.6)2=9∗16∗0.50.36=200 VV = \frac{9 * 10^9 * 16 * 10^{-9} * \cos 60^\circ}{(0.6)^2} = \frac{9 * 16 * 0.5}{0.36} = 200\ V. Chapter: Electrostatic Potential and Capacitance.

How to practise Electrostatic Potential and Capacitance

Chapter by chapter, against real past-paper questions, on a screen: NEET is computer based from 2027. Read the chapter once, then attempt its PYQs until your accuracy stops climbing, and let the wrong answers tell you which lines to reread. If the chapter keeps refusing to stick, that has its own guide. There is also a free NEET mini mock with answers.

Nearby Physics chapters

Questions

How many questions did Electrostatic Potential and Capacitance have in NEET 2026?

Electrostatic Potential and Capacitance carried 6 questions in the NEET 2026 paper. Weightage shifts between cycles, so treat it as a probability, not a promise.

Where can I practise Electrostatic Potential and Capacitance PYQs for NEET?

Explore available chapter-wise NEET practice in BrainIt, free until the 2027 exam. Review explanations and track chapter accuracy to guide your revision. Sample questions from this chapter are on this page.

Is Electrostatic Potential and Capacitance important for NEET 2027?

It is part of the Physics syllabus the paper draws from, and NEET 2027 is computer based, so practise it on a screen. Your own accuracy in the chapter matters more than any weightage table: a heavy chapter you already score well on needs less time than a light one you keep missing.

Something here wrong or out of date? Email hello@brainit.pro and a person will read it.