JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Electrostatics

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A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is

100 V / m

, then the value of Q is

____

nC.
(

ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} / Nm^{2}

and

π = 3.14

)

(A)2.14✓ Correct
(B)2.44
(C)3.25
(D)0.7

Answer: (A)

Solution

The field at the centre of a uniformly charged semicircular ring is

E = \frac{2 k λ}{R}

with

λ = \frac{Q}{π R}

, hence

E = \frac{2 k Q}{π R ^{2}}

. Solving for the charge,

Q = \frac{E π R ^{2}}{2 k} = \frac{100 \times 3.14 \times ( 0.35 ) ^{2}}{2 \times 9 \times 1 0 ^{9}} = 2.14 \times 1 0^{- 9}

C. The charge is

2.14

nC.

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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