Before You Begin

Question #42842

Difficulty: 8/10 | 🟢 Free

The Question

An isolated thin metallic spherical shell of initial radius $R_0 \text{ m}$ carries a constant total charge $Q \text{ C}$ . The shell is in a vacuum. Initially, the electric field magnitude just outside the surface of the shell is $E_0 \text{ V/m}$ and the total electric flux through a spherical Gaussian surface infinitesimally larger than the shell is $\Phi_0 \text{ V}\cdot\text{m}$ . The temperature of the shell is then uniformly increased by $\Delta T \text{ K}$ . If the coefficient of linear thermal expansion of the shell material is $\alpha \text{ K}^{-1}$ (where $\alpha \Delta T \ll 1$ ), what are the fractional changes in the electric field magnitude ( $\frac{\Delta E}{E_0}$ ) just outside the shell and the total electric flux ( $\frac{\Delta \Phi}{\Phi_0}$ ) through the new infinitesimally larger Gaussian surface, respectively?

Options

A.
$-2\alpha \Delta T$ and $0$
B.
$0$ and $2\alpha \Delta T$
C.
$2\alpha \Delta T$ and $0$
D.
$-2\alpha \Delta T$ and $2\alpha \Delta T$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Flux and Surface Geometry|Thermal Expansion of Area and Volume|Surface Integration

Why This Question Is Tricky

Common Mistakes

This question tests the conceptual distinction between Gauss's Law (which dictates total flux) and the local electric field boundary condition at a conductor's surface.

A common trap for JEE candidates is assuming that because the total enclosed charge and total flux remain constant, the electric field just outside the surface must also remain constant.

However, thermal expansion increases the surface area ( $A \propto R^2$ ), which decreases the surface charge density ( $\sigma = Q/A$ ).

Consequently, the local electric field ( $E = \sigma/\epsilon_0$ ) decreases by a fractional amount of $-2\alpha \Delta T$ , even though the total flux ( $\Phi = \oint \vec{E} \cdot d\vec{A}$ ) is perfectly conserved.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Flux and Surface Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free