JEE Main 2026 Apr 6 Shift 1, Physics Q13: Coulomb's Law And Electric Field

JEE Main2026Apr 6, Shift 1Physics

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
$2.14$
B
$2.44$
C
$3.25$
D
$0.7$

Solution

For a uniformly charged half ring of radius $R$ carrying total charge $Q$ , the field at the centre is $E = \frac{k Q}{π R ^{2} /2} \cdot \frac{2}{π}$ — equivalently $E = \frac{2 k λ}{R}$ where $λ = Q / (π R)$ is the linear charge density. Combined:

$E = \frac{2 k Q}{π R ^{2}}$ .

Solve for $Q$ : $Q = \frac{E π R ^{2}}{2 k} = \frac{100 \cdot 3.14 \cdot ( 0.35 ) ^{2}}{2 \cdot 9 \times 1 0 ^{9}}$ .

$= \frac{100 \cdot 3.14 \cdot 0.1225}{1.8 \times 1 0 ^{10}} \approx 2.14 \times 1 0^{- 9}$ C $= 2.14$ nC.

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JEE Main 2026 Apr 6 Shift 1, Physics Q13: Coulomb's Law And Electric Field

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