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MCQEasyJEE 2023Newton's Law of Gravitation

JEE Physics 2023 Question with Solution

If VV is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-

  • A

    43V\frac{4}{3}V

  • B

    VV

  • C

    3V2\frac{3V}{2}

  • D

    V2\frac{V}{2}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The potential on the surface of a sphere is VV.

Find: The potential at the center of the sphere.

For a uniformly charged non-conducting sphere, the potential inside the sphere at a distance rr from the center is given by

V(r)=14πϵ0⋅QR(32−r22R2)V(r) = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \left( \frac{3}{2} - \frac{r^2}{2R^2} \right)

At the center, r=0r = 0. Therefore,

Vcenter=14πϵ0⋅QR⋅32V_{\text{center}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \cdot \frac{3}{2}

At the surface, r=Rr = R, so

Vsurface=14πϵ0⋅QRV_{\text{surface}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R}

Hence,

Vcenter=32VsurfaceV_{\text{center}} = \frac{3}{2} V_{\text{surface}}

Since the surface potential is VV, we get

Vcenter=3V2V_{\text{center}} = \frac{3V}{2}

Therefore, the potential at the center is 3V2\frac{3V}{2}. The correct option is C. The solution labels the option as D, but the worked result matches option C in the given list.

Relation Between Center and Surface Potential

Given: Surface potential is VV.

Find: Center potential in terms of VV.

Use the standard result for a uniformly distributed spherical charge or mass distribution: the potential inside varies as

V(r)∝(32−r22R2)V(r) \propto \left( \frac{3}{2} - \frac{r^2}{2R^2} \right)

At r=Rr = R,

Vsurface∝1V_{\text{surface}} \propto 1

At r=0r = 0,

Vcenter∝32V_{\text{center}} \propto \frac{3}{2}

So the ratio is

VcenterVsurface=32\frac{V_{\text{center}}}{V_{\text{surface}}} = \frac{3}{2}

Hence,

Vcenter=3V2V_{\text{center}} = \frac{3V}{2}

Therefore, the answer is 3V2\frac{3V}{2}.

Common mistakes

  • Using the surface potential formula for every point inside the sphere is incorrect because the inside potential is not constant with position for a uniformly distributed sphere. Use the inside-potential expression and then substitute r=0r = 0 for the center.

  • Confusing a uniformly charged non-conducting sphere with a conducting sphere is wrong because for a conductor the potential is constant throughout the interior, but for a uniformly distributed sphere the center potential is larger than the surface value. Identify the physical model before applying formulas.

  • Substituting r=Rr = R instead of r=0r = 0 when asked for the center gives the surface value again. Always check which point the question asks about before evaluation.

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