JEE Challenger
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PhyMedAdv 2014 • Paper 1

Electric Fields of Point Charge Line Charge and Plane Sheet

Suppose E1(r)E_1(r), E2(r)E_2(r), and E3(r)E_3(r) denote the electric field magnitudes at a radial distance rr from a point charge QQ, an infinitely long line charge with uniform linear density λ\lambda, and an infinite planar sheet with uniform surface charge density σ\sigma, respectively. Given that E1(r0)=E2(r0)=E3(r0)E_1(r_0) = E_2(r_0) = E_3(r_0) at a specific distance r0r_0, select the correct option(s).

Options

A

Q=4σπr02Q = 4\sigma\pi r_0^2

B

r0=λ2πσr_0 = \frac{\lambda}{2\pi\sigma}

C

E1(r0/2)=2E2(r0/2)E_1(r_0/2) = 2E_2(r_0/2)

Correct
D

E2(r0/2)=4E3(r0/2)E_2(r_0/2) = 4E_3(r_0/2)

Topics & Concepts

Electric Charges and FieldsElectric FieldGauss's Law

Step-by-Step Solution

To determine the correct option(s), we write the expressions for the electric field magnitudes as functions of the radial distance rr:

  1. Electric field due to a point charge QQ: E1(r)=Q4πε0r2E_1(r) = \frac{Q}{4\pi\varepsilon_0 r^2}

  2. Electric field due to an infinitely long line charge with linear charge density λ\lambda: E2(r)=λ2πε0rE_2(r) = \frac{\lambda}{2\pi\varepsilon_0 r}

  3. Electric field due to an infinite planar sheet with uniform surface charge density σ\sigma: E3(r)=σ2ε0E_3(r) = \frac{\sigma}{2\varepsilon_0}

Given that E1(r0)=E2(r0)=E3(r0)=E0E_1(r_0) = E_2(r_0) = E_3(r_0) = E_0 at a distance r0r_0, we have: E1(r0)=Q4πε0r02=E0E_1(r_0) = \frac{Q}{4\pi\varepsilon_0 r_0^2} = E_0 E2(r0)=λ2πε0r0=E0E_2(r_0) = \frac{\lambda}{2\pi\varepsilon_0 r_0} = E_0 E3(r0)=σ2ε0=E0E_3(r_0) = \frac{\sigma}{2\varepsilon_0} = E_0

Let's evaluate each option:

  • Checking Option (A): Equating E1(r0)E_1(r_0) and E3(r0)E_3(r_0): Q4πε0r02=σ2ε0  ⟹  Q=2πσr02\frac{Q}{4\pi\varepsilon_0 r_0^2} = \frac{\sigma}{2\varepsilon_0} \implies Q = 2\pi\sigma r_0^2 Thus, Option (A) is incorrect.

  • Checking Option (B): Equating E2(r0)E_2(r_0) and E3(r0)E_3(r_0): λ2πε0r0=σ2ε0  ⟹  r0=λπσ\frac{\lambda}{2\pi\varepsilon_0 r_0} = \frac{\sigma}{2\varepsilon_0} \implies r_0 = \frac{\lambda}{\pi\sigma} Thus, Option (B) is incorrect.

  • Evaluating the fields at r=r02r = \frac{r_0}{2}: E1(r02)=Q4πε0(r02)2=4(Q4πε0r02)=4E0E_1\left(\frac{r_0}{2}\right) = \frac{Q}{4\pi\varepsilon_0 \left(\frac{r_0}{2}\right)^2} = 4 \left(\frac{Q}{4\pi\varepsilon_0 r_0^2}\right) = 4E_0

    E2(r02)=λ2πε0(r02)=2(λ2πε0r0)=2E0E_2\left(\frac{r_0}{2}\right) = \frac{\lambda}{2\pi\varepsilon_0 \left(\frac{r_0}{2}\right)} = 2 \left(\frac{\lambda}{2\pi\varepsilon_0 r_0}\right) = 2E_0

    E3(r02)=σ2ε0=E0E_3\left(\frac{r_0}{2}\right) = \frac{\sigma}{2\varepsilon_0} = E_0

  • Checking Option (C): E1(r02)=4E0E_1\left(\frac{r_0}{2}\right) = 4E_0 2E2(r02)=2×(2E0)=4E02 E_2\left(\frac{r_0}{2}\right) = 2 \times (2E_0) = 4E_0 Since E1(r02)=2E2(r02)E_1\left(\frac{r_0}{2}\right) = 2 E_2\left(\frac{r_0}{2}\right), Option (C) is correct.

  • Checking Option (D): E2(r02)=2E0E_2\left(\frac{r_0}{2}\right) = 2E_0 4E3(r02)=4×E0=4E04 E_3\left(\frac{r_0}{2}\right) = 4 \times E_0 = 4E_0 Since 2E0≠4E02E_0 \neq 4E_0, Option (D) is incorrect.

Therefore, the only correct option is C.

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