Let $E_{1}(r), E_{2}(r)$ and $E_{3}(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. If $E_{1}\left(r_{0}\right)=E_{2}\left(r_{0}\right)=E_{3}\left(r_{0}\right)$ at a given distance $r_{0}$, then

(C) $E_{1}\left(r_{0} / 2\right)=2 E_{2}\left(r_{0} / 2\right)$

(A) $Q=4 \sigma \pi r_{0}^{2}$

(B) $r_{0}=\frac{\lambda}{2 \pi \sigma}$

(D) $E_{2}\left(r_{0} / 2\right)=4 E_{3}\left(r_{0} / 2\right)$

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Electrostatics

Let

E_{1} (r), E_{2} (r)

and

E_{3} (r)

be the respective electric fields at a distance

r

from a point charge

Q

, an infinitely long wire with constant linear charge density

λ

, and an infinite plane with uniform surface charge density

σ

. If

E_{1} (r_{0}) = E_{2} (r_{0}) = E_{3} (r_{0})

at a given distance

r_{0}

, then

Answer: (C)

The fields are:

E_{1} (r) = \frac{Q}{4 π ε _{0} r ^{2}}

E_{2} (r) = \frac{λ}{2 π ε _{0} r}

E_{3} (r) = \frac{σ}{2 ε _{0}}

(constant).

From

E_{1} (r_{0}) = E_{2} (r_{0})

\frac{Q}{4 π ε _{0} r _{0}^{2}} = \frac{λ}{2 π ε _{0} r _{0}}

, giving

Q = 2 λ r_{0}

.

From

E_{1} (r_{0}) = E_{3} (r_{0})

\frac{Q}{4 π ε _{0} r _{0}^{2}} = \frac{σ}{2 ε _{0}}

, giving

Q = 2 π σ r_{0}^{2}

.

Combining:

2 λ r_{0} = 2 π σ r_{0}^{2}

, so

σ = \frac{λ}{π r _{0}}

. Then

Q = 2 π r_{0}^{2} σ

, not

4 π r_{0}^{2} σ

, so (A) is wrong. Also

r_{0} = \frac{λ}{π σ}

, not

\frac{λ}{2 π σ}

, so (B) is wrong.

At

r_{0} /2

E_{1} (r_{0} /2) = \frac{Q}{4 π ε _{0} ( r _{0} /2 ) ^{2}} = 4 E_{1} (r_{0})

and

E_{2} (r_{0} /2) = \frac{λ}{2 π ε _{0} ( r _{0} /2 )} = 2 E_{2} (r_{0})

. Since

E_{1} (r_{0}) = E_{2} (r_{0})

, we have

E_{1} (r_{0} /2) = 4 E_{1} (r_{0}) = 4 E_{2} (r_{0}) = 2 \times 2 E_{2} (r_{0}) = 2 E_{2} (r_{0} /2)

. So (C) is correct.

E_{3}

is constant, so

E_{3} (r_{0} /2) = E_{3} (r_{0}) = E_{2} (r_{0}) = \frac{1}{2} E_{2} (r_{0} /2)

, giving

E_{2} (r_{0} /2) = 2 E_{3} (r_{0} /2)

, not 4. So (D) is wrong.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options