JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Electrostatics

A thin metallic wire having cross sectional area of

1 0^{- 4} m^{2}

is used to make a ring of radius 30 cm. A positive charge of

2 π C

is uniformly distributed over the ring, while another positive charge of 30 pC is kept at the centre of the ring. The tension in the ring is ______ N; provided that the ring does not get deformed (neglect the influence of gravity). (given,

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

SI units)

Answer: 3

Solution

Take an element of the ring subtending

d θ

at the centre; its charge is

d q = \frac{Q}{2 π R} R d θ

. The radial (outward) electrostatic force from the central charge

q_{0}

on this element is

d F = \frac{k q _{0} d q}{R ^{2}} = \frac{k q _{0} Q}{2 π R ^{2}} d θ

This is balanced by the tension across the element,

2 T sin (d θ /2) \approx T d θ

, giving

T = \frac{k q _{0} Q}{2 π R ^{2}}

With

k = 9 \times 1 0^{9}

q_{0} = 30 \times 1 0^{- 12}

Q = 2 π

R = 0.3

T = \frac{9 \times 1 0 ^{9} \times 30 \times 1 0 ^{- 12} \times 2 π}{2 π \times ( 0.3 ) ^{2}} = \frac{9 \times 1 0 ^{9} \times 30 \times 1 0 ^{- 12}}{0.09} = 3 N

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer

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JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Electrostatics

Solution

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