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JEE Main Electric Charges and Fields — practice questions

Explore 141 free Electric Charges and Fields questions in JEE Main Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electric Charges and Fields: quick revision

Electric fields add as vectors. For point charges in vacuum, Coulomb’s law gives the force magnitude and the charge signs determine attraction or repulsion. The field direction is the force direction on a positive test charge.

Key ideas and when to use them

  • $F=k|q_1q_2|/r^2$, with $k=1/(4\pi\epsilon_0)$.
  • Point-charge field magnitude: $E=k|q|/r^2$.
  • Force in a field: $\vec F=q\vec E$.

Common mistakes

  • Convert microcoulombs to coulombs before calculating.
  • The force on a negative charge is opposite to the electric field.

Three worked examples

Example 1: Charges +1 μC and +2 μC are 0.30 m apart in vacuum. Use k = 9 × 10⁹ N·m²/C². Find the force magnitude.
  1. Convert each charge to coulombs.
  2. $F=(9\times10^9)(10^{-6})(2\times10^{-6})/0.30^2$.

0.20 N, repulsive

Example 2: Find the field magnitude 0.30 m from a +2 μC point charge in vacuum, using k = 9 × 10⁹.
  1. Apply the point-charge field relation.
  2. $E=(9\times10^9)(2\times10^{-6})/0.30^2$.

2 × 10⁵ N/C

Example 3: A charge −3 μC is in a uniform field of +2000 N/C along x. Find its x-component of force.
  1. Keep the negative charge sign.
  2. $F_x=(-3\times10^{-6})\times2000$.

−0.006 N

Qallery AI-assisted study notes and original examples. Further reading: NCERT Physics — Electric Charges and Fields.

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Electric Charges and Fields questions and exercises with answers

Showing questions 1–30 of 141.

A convex lens forms a real image at $60$ cm on one side of an object placed $20$ cm in front of it. Find the focal length of the lens.Read the answer and worked solution →The principle that charge can be neither created nor destroyed is:Read the answer and worked solution →A charge of 1 C placed in an electric field of 100 N/C experiences force:Read the answer and worked solution →Electric potential V due to point charge q at distance r:Read the answer and worked solution →Potential difference between two points is 12 V. Work done to move 2 C between them:Read the answer and worked solution →Electric field INSIDE a conductor in electrostatic equilibrium is:Read the answer and worked solution →At an axial point of an electric dipole at distance r >> a (a = dipole half-length, p = qa·2 the dipole moment), E equals:Read the answer and worked solution →Number of charges that can be enclosed in a Gaussian surface in vacuum:Read the answer and worked solution →For a parallel plate capacitor with plate area A and separation d, the capacitance in vacuum:Read the answer and worked solution →Capacitance changes when DIELECTRIC of relative permittivity K is inserted:Read the answer and worked solution →Energy stored in a capacitor with charge Q and capacitance C:Read the answer and worked solution →Two charges +q and +q are placed at corners of an equilateral triangle. The third corner has potential:Read the answer and worked solution →For a spherical conductor of radius R carrying charge Q, the potential AT THE SURFACE equals:Read the answer and worked solution →Equipotential surfaces around a point charge are:Read the answer and worked solution →Two capacitors 2 µF and 4 µF connected in SERIES across 12 V. Charge on each:Read the answer and worked solution →3 capacitors of 6 µF each in PARALLEL: total capacitance:Read the answer and worked solution →Two charged spheres each of radius R, charge Q, separated by distance d >> R. Force F between them is:Read the answer and worked solution →Potential at center of square (side a) with charges +q, -q, +q, -q at corners (alternating):Read the answer and worked solution →Electric field inside a spherical cavity in a uniformly charged conductor is:Read the answer and worked solution →Find work done to bring a +2 µC charge from infinity to a point where V = 50 V:Read the answer and worked solution →A charged particle q is held at rest in vertical equilibrium in a uniform electric field E pointing upward. Charge-to-mass ratio q/m:Read the answer and worked solution →In a capacitor circuit, when a dielectric is inserted while voltage is HELD CONSTANT by battery, the CHARGE on capacitor:Read the answer and worked solution →The SI unit of electric potential isRead the answer and worked solution →Electric potential at infinity in the standard syllabus convention is taken asRead the answer and worked solution →The SI unit of capacitance isRead the answer and worked solution →Electric field at any point inside a hollow charged conductor isRead the answer and worked solution →On an equipotential surface, the work done in moving a test charge between any two points isRead the answer and worked solution →The potential at a distance 0.30 m from a point charge of +20 nC is (k = 9 × 10⁹ N m² C⁻²)Read the answer and worked solution →A parallel-plate capacitor in vacuum has plate area A and separation d. Its capacitance isRead the answer and worked solution →Three capacitors of 2 µF each are connected in series. The equivalent capacitance isRead the answer and worked solution →

Worked practice questions

A convex lens forms a real image at $60$ cm on one side of an object placed $20$ cm in front of it. Find the focal length of the lens.

  1. $10$ cm
  2. $12$ cm
  3. $15$ cm
  4. $20$ cm
Show answer and solution

Correct answer: C. $15$ cm

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Sign convention: $u = -20$ cm (object on the left), $v = +60$ cm (real image on the right). We have $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20}$. That is $\dfrac{1}{f} = \dfrac{1}{60} + \dfrac{3}{60}$. That is $\dfrac{1}{f} = \dfrac{4}{60}$. So $\dfrac{1}{f} = \dfrac{1}{15}$. So $f = 15$ cm. **Trap.** Forgetting the sign on $u$ gives $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{1}{30}$, leading to a negative $f$ — that's how option B (12) or A (10) creeps in.

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The principle that charge can be neither created nor destroyed is:

  1. Conservation of energy
  2. Conservation of charge
  3. Conservation of momentum
  4. Coulomb's law
Show answer and solution

Correct answer: B. Conservation of charge

Total charge in an isolated system is conserved. Charge can be transferred between bodies but not created or destroyed. This is a fundamental conservation law in physics.

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A charge of 1 C placed in an electric field of 100 N/C experiences force:

  1. 1 N
  2. 10 N
  3. 100 N
  4. 1000 N
Show answer and solution

Correct answer: C. 100 N

We have F = qE. That is F = 1 × 100. So F = 100 N.

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