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JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2014 · Paper 1 · Physics · Electrostatics

Let E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric...

Let E1(r),E2(r){E}_{1}\left(r\right), {E}_{2}\left(r\right) and {E}_{3}\left(r\right)betherespectiveelectricfieldsatadistancerfromapointchargeQ,aninfinitelylongwirewithconstantlinearchargedensitybe the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density\lambda,andaninfiniteplanewithuniformsurfacechargedensity, and an infinite plane with uniform surface charge density\sigma.If. If{E}_{1}(r0)=E2(r0)=E3(r0)\left({r}_{0}\right) = {E}_{2}\left({r}_{0}\right) = {E}_{3}\left({r}_{0}\right) at a given distance r0{r}_{0} , then

Options

  1. AQ=4σπr02Q = 4\sigma\pi{r}_{0}^{2}
  2. Br0=λ2πσ{r}_{0} = \dfrac{\lambda}{2\pi\sigma}
  3. CE1(r02)=2E2(r02){E}_{1}\left(\dfrac{{r}_{0}}{2}\right) = 2{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)
  4. DE2(r02)=4E3(r02){E}_{2}\left(\dfrac{{r}_{0}}{2}\right) = 4{E}_{3}\left(\dfrac{{r}_{0}}{2}\right)

Correct answer

C. E1(r02)=2E2(r02){E}_{1}\left(\dfrac{{r}_{0}}{2}\right) = 2{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)

Step-by-step solution

Q4πε0r02=λ2πε0r0=σ2ε0\dfrac{Q}{4\pi{ε}_{0}{r}_{0}^{2}} = \dfrac{\lambda}{2\pi{ε}_{0}{r}_{0}} = \dfrac{\sigma}{2{ε}_{0}}
E1(r02)=Qπε0r02,E2{E}_{1}\left(\dfrac{{r}_{0}}{2}\right) = \dfrac{Q}{\pi{ε}_{0}{r}_{0}^{2}}, {E}_{2}\left(\dfrac{{r}_{0}}{2}\right) = \dfrac{\lambda}{\pi{ε}_{0}{r}_{0}}, {E}_{3}\left(\dfrac{{r}_{0}}{2}\right) = \dfrac{\sigma}{2{ε}_{0}}<br><br >{E}_{1}\left(\dfrac{{r}_{0}}{2}\right) = 2{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)
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