Suppose a uniformly charged wall provides a uniform electric field of 2 \times 10^4 ~N / C normally. A charged particle of mass 2 ~g being suspended through a silk thread of length 20 ~cm and remain stayed at a distance of 10 ~cm from the wall. Then the charge on the particle will be \frac1\sqrtx \mu C where x= \underline\hspace2cm . [use g=10 ~m / s^2 ]

Electrostatics PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q84 with Solution

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Correct answer: 3

Given data

Electric field due to wall: $E = 2 \times 10^4\,\text{N/C}$
Mass of particle: $m = 2\,\text{g} = 2 \times 10^{-3}\,\text{kg}$
Length of silk thread: $l = 20\,\text{cm} = 0.20\,\text{m}$
Distance of particle from wall: $d = 10\,\text{cm} = 0.10\,\text{m}$
$g = 10\,\text{m/s}^2$

The wall produces electric field normal to the wall, so the electric force on the charged particle is horizontal.

Find the angle made by the thread

Since the particle is at a perpendicular distance $d = 10$ cm from the wall and thread length is $l = 20$ cm,

\sin\theta = \frac{d}{l} = \frac{10}{20} = \frac{1}{2}

Hence,

\theta = 30^\circ

Therefore,

\tan\theta = \frac{1}{\sqrt{3}}

For equilibrium of the suspended charged particle

Forces on the particle:

Weight $mg$ downward
Electric force $qE$ horizontally away/towards the wall
Tension $T$ along the thread

Resolving tension:

T\cos\theta = mg

T\sin\theta = qE

Dividing,

\tan\theta = \frac{qE}{mg}

So,

q = \frac{mg\tan\theta}{E}

Substitute values

First,

mg = 2 \times 10^{-3} \times 10 = 2 \times 10^{-2}\,\text{N}

Then,

q = \frac{(2 \times 10^{-2})\left(\frac{1}{\sqrt{3}}\right)}{2 \times 10^4}

q = \frac{10^{-6}}{\sqrt{3}}\,\text{C}

Since $1\,\mu\text{C} = 10^{-6}\,\text{C}$ ,

q = \frac{1}{\sqrt{3}}\,\mu\text{C}

Compare with the form given

Given,

q = \frac{1}{\sqrt{x}}\,\mu\text{C}

So,

x = 3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

So they agree.

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View written solutionFree

Correct answer: 3

Given data

Electric field due to wall: $E = 2 \times 10^4\,\text{N/C}$
Mass of particle: $m = 2\,\text{g} = 2 \times 10^{-3}\,\text{kg}$
Length of silk thread: $l = 20\,\text{cm} = 0.20\,\text{m}$
Distance of particle from wall: $d = 10\,\text{cm} = 0.10\,\text{m}$
$g = 10\,\text{m/s}^2$

The wall produces electric field normal to the wall, so the electric force on the charged particle is horizontal.

Find the angle made by the thread

Since the particle is at a perpendicular distance $d = 10$ cm from the wall and thread length is $l = 20$ cm,

\sin\theta = \frac{d}{l} = \frac{10}{20} = \frac{1}{2}

Hence,

\theta = 30^\circ

Therefore,

\tan\theta = \frac{1}{\sqrt{3}}

For equilibrium of the suspended charged particle

Forces on the particle:

Weight $mg$ downward
Electric force $qE$ horizontally away/towards the wall
Tension $T$ along the thread

Resolving tension:

T\cos\theta = mg

T\sin\theta = qE

Dividing,

\tan\theta = \frac{qE}{mg}

So,

q = \frac{mg\tan\theta}{E}

Substitute values

First,

mg = 2 \times 10^{-3} \times 10 = 2 \times 10^{-2}\,\text{N}

Then,

q = \frac{(2 \times 10^{-2})\left(\frac{1}{\sqrt{3}}\right)}{2 \times 10^4}

q = \frac{10^{-6}}{\sqrt{3}}\,\text{C}

Since $1\,\mu\text{C} = 10^{-6}\,\text{C}$ ,

q = \frac{1}{\sqrt{3}}\,\mu\text{C}

Compare with the form given

Given,

q = \frac{1}{\sqrt{x}}\,\mu\text{C}

So,

x = 3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

So they agree.

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