Figure shows a rod AB, which is bent in a 120 ^\circ circular arc of radius R. A charge ( - Q) is uniformly distributed over rod AB. What is the electric field \overrightarrow E at the centre of curvature O ?

3\sqrt 3 Q \over 8\pi ^2\varepsilon _0R^2(\widehat i)

3\sqrt 3 Q \over 8\pi \varepsilon _0R^2(\widehat i)

3\sqrt 3 Q \over 16\pi ^2\varepsilon _0R^2(\widehat i)

3\sqrt 3 Q \over 8\pi ^2\varepsilon _0R^2( - \widehat i)

Electrostatics PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q58 with Solution

View written solutionFree

Correct answer: B

Geometry of the arc

The rod is a circular arc of radius $R$ subtending an angle $120^\circ = \dfrac{2\pi}{3}$ at the centre $O$ .

It is symmetric about the $x$ -axis, so the vertical components of electric field at $O$ cancel out. Hence the net field will lie along the horizontal axis.

Since the rod carries negative charge, the electric field at the centre points towards the arc. From the figure/options, that is along $+\hat i$ .

Linear charge density

Total charge on the arc is $-Q$ uniformly distributed.

Arc length:

L = R\theta = R\left(\frac{2\pi}{3}\right)

So the magnitude of linear charge density is

\lambda = \frac{Q}{L} = \frac{Q}{(2\pi R/3)} = \frac{3Q}{2\pi R}

(Here I use $Q$ as magnitude; direction is handled separately.)

Field due to a small element

Take a small charge element at angle $\phi$ from the symmetry axis.

Then

dq = \lambda R\, d\phi

Magnitude of electric field at $O$ due to this element:

dE = \frac{1}{4\pi\varepsilon_0}\frac{dq}{R^2}

Only the horizontal component survives after symmetry:

dE_x = dE\cos\phi

Thus

dE_x = \frac{1}{4\pi\varepsilon_0}\frac{dq}{R^2}\cos\phi

Substitute $dq = \lambda R\, d\phi$ :

dE_x = \frac{1}{4\pi\varepsilon_0}\frac{\lambda R\, d\phi}{R^2}\cos\phi = \frac{\lambda}{4\pi\varepsilon_0 R}\cos\phi\, d\phi

Integrate over the arc

The arc subtends $120^\circ$ , so relative to the symmetry axis the limits are

\phi = -60^\circ \text{ to } +60^\circ

that is,

\phi = -\frac{\pi}{3} \text{ to } \frac{\pi}{3}

Therefore

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\int_{-\pi/3}^{\pi/3} \cos\phi\, d\phi

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left[\sin\phi\right]_{-\pi/3}^{\pi/3}

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left(\sin\frac{\pi}{3}-\sin\left(-\frac{\pi}{3}\right)\right)

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right) = \frac{\lambda\sqrt3}{4\pi\varepsilon_0 R}

Now substitute

\lambda = \frac{3Q}{2\pi R}

E_x = \frac{\sqrt3}{4\pi\varepsilon_0 R}\cdot \frac{3Q}{2\pi R} = \frac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}

Hence,

\boxed{\vec E = \frac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}\,\hat i}

Option check

A: Missing one factor of $\pi$ in denominator, so incorrect.
B: $\dfrac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}\hat i$ correct.
C: Extra factor $2$ in denominator, incorrect.
D: Correct magnitude but wrong direction.

Therefore the correct option is

\boxed{\text{B}}

More from Electrostatics

View written solutionFree

Correct answer: B

Geometry of the arc

The rod is a circular arc of radius $R$ subtending an angle $120^\circ = \dfrac{2\pi}{3}$ at the centre $O$ .

It is symmetric about the $x$ -axis, so the vertical components of electric field at $O$ cancel out. Hence the net field will lie along the horizontal axis.

Since the rod carries negative charge, the electric field at the centre points towards the arc. From the figure/options, that is along $+\hat i$ .

Linear charge density

Total charge on the arc is $-Q$ uniformly distributed.

Arc length:

L = R\theta = R\left(\frac{2\pi}{3}\right)

So the magnitude of linear charge density is

\lambda = \frac{Q}{L} = \frac{Q}{(2\pi R/3)} = \frac{3Q}{2\pi R}

(Here I use $Q$ as magnitude; direction is handled separately.)

Field due to a small element

Take a small charge element at angle $\phi$ from the symmetry axis.

Then

dq = \lambda R\, d\phi

Magnitude of electric field at $O$ due to this element:

dE = \frac{1}{4\pi\varepsilon_0}\frac{dq}{R^2}

Only the horizontal component survives after symmetry:

dE_x = dE\cos\phi

Thus

dE_x = \frac{1}{4\pi\varepsilon_0}\frac{dq}{R^2}\cos\phi

Substitute $dq = \lambda R\, d\phi$ :

dE_x = \frac{1}{4\pi\varepsilon_0}\frac{\lambda R\, d\phi}{R^2}\cos\phi = \frac{\lambda}{4\pi\varepsilon_0 R}\cos\phi\, d\phi

Integrate over the arc

The arc subtends $120^\circ$ , so relative to the symmetry axis the limits are

\phi = -60^\circ \text{ to } +60^\circ

that is,

\phi = -\frac{\pi}{3} \text{ to } \frac{\pi}{3}

Therefore

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\int_{-\pi/3}^{\pi/3} \cos\phi\, d\phi

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left[\sin\phi\right]_{-\pi/3}^{\pi/3}

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left(\sin\frac{\pi}{3}-\sin\left(-\frac{\pi}{3}\right)\right)

E_x = \frac{\lambda}{4\pi\varepsilon_0 R}\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right) = \frac{\lambda\sqrt3}{4\pi\varepsilon_0 R}

Now substitute

\lambda = \frac{3Q}{2\pi R}

E_x = \frac{\sqrt3}{4\pi\varepsilon_0 R}\cdot \frac{3Q}{2\pi R} = \frac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}

Hence,

\boxed{\vec E = \frac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}\,\hat i}

Option check

A: Missing one factor of $\pi$ in denominator, so incorrect.
B: $\dfrac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}\hat i$ correct.
C: Extra factor $2$ in denominator, incorrect.
D: Correct magnitude but wrong direction.

Therefore the correct option is

\boxed{\text{B}}

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