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Electrostatics question

2020 · 8 Jan · Shift 2 · Q40
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Electrostatics question

2020 · 8 Jan · Shift 2 · Q40

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2 = R1/R2. Then the ratio V1 (on S1) / V2 (on S2) of the electrostatic potentials on each sphere is :
  1. A
    (R1/R2)2
  2. B
    (R2/R1)
  3. C
    (R1/R2)3
  4. D
    R1/R2
View written solutionFree

Correct answer: A

  1. For an isolated charged metallic sphere of radius RRR and charge QQQ:

    • Electric field at the surface: E=14πε0QR2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}E=4πε0​1​R2Q​
    • Potential at the surface: V=14πε0QRV = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}V=4πε0​1​RQ​

  2. Relate EEE and VVV for a sphere: E=VR⇒V=ERE = \frac{V}{R} \quad \Rightarrow \quad V = ERE=RV​⇒V=ER

  3. Apply this to the two spheres: V1=E1R1,V2=E2R2V_1 = E_1 R_1, \qquad V_2 = E_2 R_2V1​=E1​R1​,V2​=E2​R2​

  4. Therefore, V1V2=E1R1E2R2=E1E2⋅R1R2\frac{V_1}{V_2} = \frac{E_1R_1}{E_2R_2} = \frac{E_1}{E_2}\cdot \frac{R_1}{R_2}V2​V1​​=E2​R2​E1​R1​​=E2​E1​​⋅R2​R1​​

  5. Given: E1E2=R1R2\frac{E_1}{E_2} = \frac{R_1}{R_2}E2​E1​​=R2​R1​​ Substituting, V1V2=R1R2⋅R1R2=(R1R2)2\frac{V_1}{V_2} = \frac{R_1}{R_2}\cdot \frac{R_1}{R_2} = \left(\frac{R_1}{R_2}\right)^2V2​V1​​=R2​R1​​⋅R2​R1​​=(R2​R1​​)2

  6. Check options:

    • A: (R1R2)2\left(\dfrac{R_1}{R_2}\right)^2(R2​R1​​)2 ✅
    • B: R2R1\dfrac{R_2}{R_1}R1​R2​​ ❌
    • C: (R1R2)3\left(\dfrac{R_1}{R_2}\right)^3(R2​R1​​)3 ❌
    • D: R1R2\dfrac{R_1}{R_2}R2​R1​​ ❌

Hence, the correct answer is Option A.

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