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Electrostatics question

2024 · 30 Jan · Shift 2 · Q75

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2024 · 30 Jan · Shift 2 · Q75

Electrostatics question

2024 · 30 Jan · Shift 2 · Q75

JEE MainPhysicsElectrostaticsMCQ+4 / −1

A particle of charge '

-q

' and mass '

m

' moves in a circle of radius '

r

' around an infinitely long line charge of linear charge density '

+\lambda

'. Then time period will be given as : (Consider

k

as Coulomb's constant)

A
$T^2=\frac{4 \pi^2 m}{2 k \lambda q} r^3$
B
$T=\frac{1}{2 \pi r} \sqrt{\frac{m}{2 k \lambda q}}$
C
$T=\frac{1}{2 \pi} \sqrt{\frac{2 k \lambda q}{m}}$
D
$T=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$

View written solutionFree

Correct answer: D

Electric field due to an infinite line charge

For an infinitely long line charge of linear charge density $+\lambda$ , the electric field at distance $r$ is

E = \frac{2k\lambda}{r}.

Electrostatic force on the particle

The particle has charge $-q$ . Its magnitude of force due to the line charge is

F = qE = q\left(\frac{2k\lambda}{r}\right)=\frac{2k\lambda q}{r}.

Since the charge is negative and the line charge is positive, the force is attractive, so it can provide the required centripetal force.

Use centripetal force condition

For circular motion of radius $r$ and speed $v$ ,

\frac{mv^2}{r} = \frac{2k\lambda q}{r}.

Cancelling $r$ from both sides,

mv^2 = 2k\lambda q

so

v = \sqrt{\frac{2k\lambda q}{m}}.

Find the time period

Time period of circular motion is

T = \frac{2\pi r}{v}.

Substitute $v$ :

T = \frac{2\pi r}{\sqrt{\frac{2k\lambda q}{m}}} = 2\pi r\sqrt{\frac{m}{2k\lambda q}}.

Match with options

Thus,

\boxed{T = 2\pi r\sqrt{\frac{m}{2k\lambda q}}}

This matches Option D.

Verification with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

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