A thin metallic wire having cross sectional area of 10^-4 ~m^2 is used to make a ring of radius 30 ~cm . A positive charge of 2 \pi ~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 \underline\hspace2cm N ; provided that the ring does not get deformed (neglect the influence of gravity). (given, \frac14 \pi \epsilon_0=9 \times 10^9 SI units)

Electrostatics PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q82 with Solution

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Correct answer: 3

Given data

Cross-sectional area of wire: $A = 10^{-4}\,\text{m}^2$
Radius of ring: $R = 30\,\text{cm} = 0.3\,\text{m}$
Charge uniformly distributed on ring: $Q = 2\pi\,\text{C}$
Charge at centre: $q = 30\,\text{pC} = 30\times 10^{-12}\,\text{C}$
Coulomb constant: $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times 10^9$

We need the tension in the ring.

Force on a small element of the ring due to the central charge

Let the linear charge density on the ring be

\lambda = \frac{Q}{2\pi R}

A small element subtending angle $d\theta$ has length

dl = R\,d\theta

and charge

dq = \lambda\,dl = \lambda R\,d\theta = \frac{Q}{2\pi R}R\,d\theta = \frac{Q}{2\pi}d\theta

The central charge $q$ repels this element radially outward with force

dF = \frac{kq\,dq}{R^2}

So,

dF = \frac{kq}{R^2}\cdot \frac{Q}{2\pi}d\theta

Relating this force to tension

Consider the small element of the ring subtending angle $d\theta$ .

If the tension in the ring is $T$ , then the two tensions at its ends have resultant radial inward magnitude

2T\sin\frac{d\theta}{2}

For small $d\theta$ ,

\sin\frac{d\theta}{2} \approx \frac{d\theta}{2}

Hence inward force due to tension is

2T\cdot \frac{d\theta}{2} = T\,d\theta

For equilibrium of the element,

T\,d\theta = dF

Therefore,

T\,d\theta = \frac{kq}{R^2}\cdot \frac{Q}{2\pi}d\theta

Cancelling $d\theta$ ,

T = \frac{kqQ}{2\pi R^2}

Substitute values

T = \frac{(9\times 10^9)(30\times 10^{-12})(2\pi)}{2\pi(0.3)^2}

The $2\pi$ cancels:

T = \frac{(9\times 10^9)(30\times 10^{-12})}{0.09}

Now,

9\times 30 = 270, \qquad 10^9\times 10^{-12} = 10^{-3}

So,

(9\times 10^9)(30\times 10^{-12}) = 270\times 10^{-3} = 0.27

Thus,

T = \frac{0.27}{0.09} = 3\,\text{N}

Final answer

The tension in the ring is

\boxed{3\,\text{N}}

So the required integer is

\boxed{3}

Comparison with stored answer

Stored correct answer: $3$

Our derived answer is also $3$ , so they agree.

More from Electrostatics

View written solutionFree

Correct answer: 3

Given data

Cross-sectional area of wire: $A = 10^{-4}\,\text{m}^2$
Radius of ring: $R = 30\,\text{cm} = 0.3\,\text{m}$
Charge uniformly distributed on ring: $Q = 2\pi\,\text{C}$
Charge at centre: $q = 30\,\text{pC} = 30\times 10^{-12}\,\text{C}$
Coulomb constant: $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times 10^9$

We need the tension in the ring.

Force on a small element of the ring due to the central charge

Let the linear charge density on the ring be

\lambda = \frac{Q}{2\pi R}

A small element subtending angle $d\theta$ has length

dl = R\,d\theta

and charge

dq = \lambda\,dl = \lambda R\,d\theta = \frac{Q}{2\pi R}R\,d\theta = \frac{Q}{2\pi}d\theta

The central charge $q$ repels this element radially outward with force

dF = \frac{kq\,dq}{R^2}

So,

dF = \frac{kq}{R^2}\cdot \frac{Q}{2\pi}d\theta

Relating this force to tension

Consider the small element of the ring subtending angle $d\theta$ .

If the tension in the ring is $T$ , then the two tensions at its ends have resultant radial inward magnitude

2T\sin\frac{d\theta}{2}

For small $d\theta$ ,

\sin\frac{d\theta}{2} \approx \frac{d\theta}{2}

Hence inward force due to tension is

2T\cdot \frac{d\theta}{2} = T\,d\theta

For equilibrium of the element,

T\,d\theta = dF

Therefore,

T\,d\theta = \frac{kq}{R^2}\cdot \frac{Q}{2\pi}d\theta

Cancelling $d\theta$ ,

T = \frac{kqQ}{2\pi R^2}

Substitute values

T = \frac{(9\times 10^9)(30\times 10^{-12})(2\pi)}{2\pi(0.3)^2}

The $2\pi$ cancels:

T = \frac{(9\times 10^9)(30\times 10^{-12})}{0.09}

Now,

9\times 30 = 270, \qquad 10^9\times 10^{-12} = 10^{-3}

So,

(9\times 10^9)(30\times 10^{-12}) = 270\times 10^{-3} = 0.27

Thus,

T = \frac{0.27}{0.09} = 3\,\text{N}

Final answer

The tension in the ring is

\boxed{3\,\text{N}}

So the required integer is

\boxed{3}

Comparison with stored answer

Stored correct answer: $3$

Our derived answer is also $3$ , so they agree.

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