JEE Advanced 2014 · Paper 1 · Physics Q5

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PhysicsMultiple correct+3Question 5 of 60

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 $E_1(r_0)=E_2(r_0)=E_3(r_0)$ at a given distance $r_0$ , then

Q=4\sigma\pi r_0^2

r_0=\dfrac{\lambda}{2\pi\sigma}

E_1(r_0/2)=2E_2(r_0/2)

E_2(r_0/2)=4E_3(r_0/2)

Answer and worked solution

Official answer: C

Fields.
$E_1=\frac{Q}{4\pi\varepsilon_0r^2},\qquad E_2=\frac{\lambda}{2\pi\varepsilon_0r},\qquad E_3=\frac{\sigma}{2\varepsilon_0}.$

(A) $E_1(r_0)=E_3(r_0)$ gives $Q=2\pi\sigma r_0^2$ , not $4\pi\sigma r_0^2$ . False.
(B) $E_2(r_0)=E_3(r_0)$ gives $r_0=\dfrac{\lambda}{\pi\sigma}$ . False.
(C) Halving $r$ multiplies $E_1$ by 4 and $E_2$ by 2. Since they were equal at $r_0$ , $E_1(r_0/2)=2E_2(r_0/2)$ . True.
(D) $E_2(r_0/2)=2E_2(r_0)=2E_3$ , and $E_3$ does not depend on distance. So the ratio is 2, not 4. False.

(C)

Topics:Electric flux and Gauss's law Electric field

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JEE Advanced 2014 · Paper 1, Physics question 5