Electrostatic Complete NEET PYQs (1988–2026)

Solved Electrostatic Complete PYQs

3 recent questions with answers. The full set of 131 is free to practise.

1. NEET NAEasyElectrostatic Potential
   Q1

   A conducting sphere of radius $R$ is given a charge $Q$. The electric potential and the electric field at the centre of the sphere respectively are
   (2014)

   • Azero and $\frac{Q}{4\pi {\epsilon }_0{R}^2}$
   • B$\frac{Q}{4\pi {\epsilon }_{0}R}$ and zero
   • C$\frac{Q}{4\pi {\epsilon }_{0}R}$ and $\frac{Q}{4\pi {\epsilon }_0{R}^2}$
   • Dboth are zero

   Show answer & solution

   Correct answer: (B)

   For the conducting sphere,
   Potential at the centre $=$ Potential on the sphere $=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$
   Electric field at the centre $=0$
2. NEET 2026HardElectrostatic Potential Energy
   Q2

   Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. A point charge $-q$ $(q<<Q)$ with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: (${ϵ}_0$ is the permittivity of vacuum, neglect gravitational
   
   forces).

   • A$\sqrt{\frac{3Qq}{4\pi {ϵ}_{0}mR}}$
   • B$\sqrt{\frac{2Qq}{3\pi {ϵ}_{0}mR}}$
   • C$\sqrt{\frac{Qq}{3\pi {ϵ}_{0}mR}}$
   • D$\sqrt{\frac{Qq}{4\pi {ϵ}_{0}mR}}$

   Show answer & solution

   Correct answer: (C)

   Potential at distance $r$ from centre of uniformly charged sphere
   
   (outside): $V=\frac{Q}{4\pi {ϵ}_{0}r}$.
   
   At $r=3R$:
   
   $V_i=\frac{Q}{4\pi {ϵ}_0(3R)}=\frac{Q}{12\pi {ϵ}_{0}R}$.
   
   At $r=R$ (surface): $V_f=\frac{Q}{4\pi {ϵ}_{0}R}$.
   
   Work done by electric field =
   
   $-q\left(V_f-V_i\right)=-q\left(\frac{Q}{4\pi {ϵ}_{0}R}-\frac{Q}{12\pi {ϵ}_{0}R}\right)=-q\left(\frac{3Q-Q}{12\pi {ϵ}_{0}R}\right)=-q\left(\frac{2Q}{12\pi {ϵ}_{0}R}\right)=-\frac{qQ}{6\pi {ϵ}_{0}R}$.
   
   This work equals increase in kinetic energy:
   
   $\frac{1}{2}m{v}^2=\frac{qQ}{6\pi {ϵ}_{0}R}$.
   
   $v=\sqrt{\frac{2qQ}{6\pi {ϵ}_{0}mR}}=\sqrt{\frac{qQ}{3\pi {ϵ}_{0}mR}}$.
3. NEET 2026EasyEnergy Stored in a Capacitor and Dielectrics
   Q3

   Consider two uncharged capacitors of equal capacitance $200pF$. One of them is charged by a $100\text{V}$ supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

   • A$0.5\times 10^{-6}J$
   • B$1.0J$
   • C$1.0\times 10^{-6}J$
   • D$0.5J$

   Show answer & solution

   Correct answer: (A)

   Energy loss
   
   $$=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2=\frac{1}{2}\left(\frac{200\times 200}{400}\right)\times 10^{-12}\times (100{)}^2$$
   
   $=\frac{1}{2}\times 100\times 10^{-12}\times 10^4=0.5\times 10^{-6}$J.

Topic-wise question count

Where NEET has asked from Electrostatic Complete, most frequent first.

• Energy Stored in a Capacitor and Dielectrics21
• Electrostatic Potential18
• Electric Field and Electric Field Lines17
• Electric Dipole – Field, Torque and Potential Energy14
• Gauss’s Law and its Applications14
• Capacitance and Capacitor11
• Electrostatic Potential Energy10
• Combination of Capacitors10
• Coulomb’s Law10
• Potential due Dipole2
• Equipotential Surfaces2
• Gauss's Law and its Applications1
• Electric Charge – Properties and Conservation1

Year-wise Electrostatic Complete questions in NEET

Number of questions from this chapter in each year's paper(s).