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Physics · chapter PYQs

Electric Charges and Fields, on the exam screen.

2 questions from Re-NEET 2026, NEET 2020, 2 minutes, marked +4 / −1. See them with solutions.

Electric Charges and Fields · PYQs

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  1. Question 1 · Re-NEET 2026 · Q33 · +4 / −1

    A point charge QQ is placed inside a cavity within a solid isolated conducting sphere. Consider points AA, BB and CC as shown in the figure, where the magnitudes of the electric fields are EAE_A, EBE_B, ECE_C, respectively. The points BB and CC are at the same distance from the center of the solid sphere. The correct option is :

    figure
    Answer and explanation

    Answer: (2)

    Point AA lies in the empty cavity, not in the conducting material, so the field of QQ is fully present there and EA≠0E_A \neq 0. The induced −Q-Q on the cavity wall and +Q+Q on the outer surface arrange themselves so that the outer surface charge is UNIFORM regardless of where the cavity sits, so outside the sphere the field is exactly that of a point charge QQ at the centre. Since BB and CC are equidistant from that centre, EB=ECE_B = E_C. NCERT Class 12, Electric Charges and Fields (electrostatics of conductors).

    Why the other options are wrong:

    - (1) confuses the cavity with the conducting material - the field vanishes inside the METAL, not inside the hollow
    - (3) combines the cavity error with the idea that the off-centre cavity biases the external field towards BB
    - (4) correctly sees EA≠0E_A \neq 0 but wrongly assumes the external field is skewed away from the cavity side

  2. Question 2 · NEET 2020 · Q170 · +4 / −1

    A spherical conductor of radius 10 cm has a charge of 3.2 × 10⁻⁷ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? (1/(4πε₀) = 9 × 10⁹ N m² C⁻²)

    Answer and explanation

    Answer: (1)

    The electric field outside a charged spherical conductor is given by E=14πϵ0Qr2E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}. Substituting the values, E=9×109×3.2×10−7(0.15)2=1.28×105 N/CE = 9 \times 10^9 \times \frac{3.2 \times 10^{-7}}{(0.15)^2} = 1.28 \times 10^5\ N/C. Chapter: Electric Charges and Fields.