JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Electrostatics

Electric charge is transferred to an irregular metallic disk as shown in figure. If

σ_{1}, σ_{2}, σ_{3}

and

σ_{4}

are charge densities at given points then, choose the correct answer from the options given below:
(A)

σ_{1} > σ_{3}; σ_{2} = σ_{4}

(B)

σ_{1} > σ_{2}; σ_{3} > σ_{4}

(C)

σ_{1} > σ_{3} > σ_{2} = σ_{4}

(D)

σ_{1} < σ_{3} < σ_{2} = σ_{4}

(E)

σ_{1} = σ_{2} = σ_{3} = σ_{4}

(A)A, B and C Only✓ Correct
(B)A and C Only
(C)D and E Only
(D)B and C Only

Answer: (A)

Solution

On a charged conductor the surface charge density is largest where the surface is most sharply curved, since

σ \propto \frac{1}{radius of curvature}

. The two pointed ends carry the highest densities, with the upper tip sharper than the lower, while the two flatter flanks share the largest radius and hence equal, smaller densities, giving

σ_{1} > σ_{3} > σ_{2} = σ_{4}

. This ordering makes the relations

σ_{1} > σ_{3}

with

σ_{2} = σ_{4}

, then

σ_{1} > σ_{2}

with

σ_{3} > σ_{4}

, and

σ_{1} > σ_{3} > σ_{2} = σ_{4}

all hold, while the other two fail. Only the first three statements are correct.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Electrostatics →

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Electrostatics

Solution

More JEE Main questions on Electrostatics