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Electrostatics question

2014 · Shift 1 · Q43
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Electrostatics question

2014 · Shift 1 · Q43

JEE AdvancedPhysicsElectrostaticsMultiple correct+3 / −1
Let E1(r),E2(r){E_1}\left( r \right),{E_2}\left( r \right)E1​(r),E2​(r) and E3(r){E_3}\left( r \right)E3​(r) be the respective electric field at a distance rrr from a point charge Q,Q,Q, an infinitely long wire with constant linear charge density λ,\lambda ,λ, and an infinite plane with uniform surface charge density σ.\sigma .σ. If E1(r0)=E2(r0)=E3(r0)E{}_1\left( {{r_0}} \right) = {E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)E1​(r0​)=E2​(r0​)=E3​(r0​) at a given distance r0.{r_0}.r0​. then
  1. A
    Q=4σπr02Q = 4\sigma \pi r_0^2Q=4σπr02​
  2. B
    r0=λ2πσ{r_0} = {\lambda \over {2\pi \sigma }}r0​=2πσλ​
  3. C
    E1(r0/2)=2E2(r0/2){E_1}\left( {{r_0}/2} \right) = 2{E_2}\left( {{r_0}/2} \right)E1​(r0​/2)=2E2​(r0​/2)
  4. D
    E2(r0/2)=4E3(r0/2){E_2}\left( {{r_0}/2} \right) = 4{E_3}\left( {{r_0}/2} \right)E2​(r0​/2)=4E3​(r0​/2)
View written solutionFree

Correct answer: C

  1. Write the electric fields

For the three charge distributions:

  • Point charge QQQ: E1(r)=14πε0Qr2E_1(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}E1​(r)=4πε0​1​r2Q​

  • Infinitely long wire with linear charge density λ\lambdaλ: E2(r)=λ2πε0rE_2(r)=\frac{\lambda}{2\pi\varepsilon_0 r}E2​(r)=2πε0​rλ​

  • Infinite plane sheet with surface charge density σ\sigmaσ: E3(r)=σ2ε0E_3(r)=\frac{\sigma}{2\varepsilon_0}E3​(r)=2ε0​σ​

  1. Use the condition at r=r0r=r_0r=r0​

Given: E1(r0)=E2(r0)=E3(r0)E_1(r_0)=E_2(r_0)=E_3(r_0)E1​(r0​)=E2​(r0​)=E3​(r0​)

So, 14πε0Qr02=λ2πε0r0=σ2ε0\frac{1}{4\pi\varepsilon_0}\frac{Q}{r_0^2}=\frac{\lambda}{2\pi\varepsilon_0 r_0}=\frac{\sigma}{2\varepsilon_0}4πε0​1​r02​Q​=2πε0​r0​λ​=2ε0​σ​

We now compare them pairwise.

  1. Relation between QQQ and σ\sigmaσ

From E1(r0)=E3(r0)E_1(r_0)=E_3(r_0)E1​(r0​)=E3​(r0​): 14πε0Qr02=σ2ε0\frac{1}{4\pi\varepsilon_0}\frac{Q}{r_0^2}=\frac{\sigma}{2\varepsilon_0}4πε0​1​r02​Q​=2ε0​σ​

Multiply both sides by 4πε0r024\pi\varepsilon_0 r_0^24πε0​r02​: Q=2πσr02Q=2\pi\sigma r_0^2Q=2πσr02​

Option A says: Q=4πσr02Q=4\pi\sigma r_0^2Q=4πσr02​

This is incorrect.

  1. Relation between r0r_0r0​, λ\lambdaλ, and σ\sigmaσ

From E2(r0)=E3(r0)E_2(r_0)=E_3(r_0)E2​(r0​)=E3​(r0​): λ2πε0r0=σ2ε0\frac{\lambda}{2\pi\varepsilon_0 r_0}=\frac{\sigma}{2\varepsilon_0}2πε0​r0​λ​=2ε0​σ​

Multiply both sides by 2ε02\varepsilon_02ε0​: λπr0=σ\frac{\lambda}{\pi r_0}=\sigmaπr0​λ​=σ

So, r0=λπσr_0=\frac{\lambda}{\pi\sigma}r0​=πσλ​

Option B says: r0=λ2πσr_0=\frac{\lambda}{2\pi\sigma}r0​=2πσλ​

So B is incorrect.

  1. Check option C

Compute fields at r=r02r=\dfrac{r_0}{2}r=2r0​​.

For point charge:

=\frac{1}{4\pi\varepsilon_0}\frac{4Q}{r_0^2}=4E_1(r_0)$$ For line charge: $$E_2\left(\frac{r_0}{2}\right)=\frac{\lambda}{2\pi\varepsilon_0 (r_0/2)}=2E_2(r_0)$$ Since $E_1(r_0)=E_2(r_0)$, $$E_1\left(\frac{r_0}{2}\right)=4E_1(r_0)=4E_2(r_0)$$ $$2E_2\left(\frac{r_0}{2}\right)=2\cdot 2E_2(r_0)=4E_2(r_0)$$ Hence, $$E_1\left(\frac{r_0}{2}\right)=2E_2\left(\frac{r_0}{2}\right)$$ So **C is correct**. --- 6. **Check option D** For the line charge: $$E_2\left(\frac{r_0}{2}\right)=2E_2(r_0)$$ For the plane sheet, field is independent of distance: $$E_3\left(\frac{r_0}{2}\right)=E_3(r_0)$$ Since $E_2(r_0)=E_3(r_0)$, $$E_2\left(\frac{r_0}{2}\right)=2E_3(r_0)=2E_3\left(\frac{r_0}{2}\right)$$ Option D says: $$E_2\left(\frac{r_0}{2}\right)=4E_3\left(\frac{r_0}{2}\right)$$ This is **incorrect**. --- 7. **Final conclusion** Only option **C** is correct.
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