HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q42 Solution: A metal sphere of radius R is charged to a potential V. 1. Find …

QPages 165-170: Q42: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q42 Solution: A metal sphere of radius R is charged to a potential V. 1. Find …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often use $U= frac12 CV^2$ to compute the energy inside a finite region, assuming it equals the total field energy. This gives $U=rac{Q^2}{8pivarepsilon_0R}$, twice the correct value. The correct approach is to integrate the energy density $u= frac12varepsilon_0E^2$ over the specific volume (here $Rle rle2R$), yielding $U_{ ext{in}}=rac{Q^2}{16pivarepsilon_0R}$. Always integrate over the region of interest rather than applying the global capacitor formula.

A metal sphere of radius R is charged to a potential V. 1. Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2R. 2. Show that the electrostatic field energy stored outside the sphere of radius 2R equals that stored within it.

Step-by-Step Verified Solution

**Answer:** (a)

U = \frac{Q^2}{16\pi\epsilon_0 R}

(b) Both are equal. **Step-by-Step Explanation:** (a) Energy within sphere of radius 2R:

U_{in} = \int_R^{2R} \frac{\epsilon_0 E^2}{2} 4\pi r^2 dr = \frac{Q^2}{16\pi\epsilon_0 R}

(b) Energy outside 2R:

U_{out} = \int_{2R}^\infty \frac{\epsilon_0 E^2}{2} 4\pi r^2 dr = \frac{Q^2}{16\pi\epsilon_0 R}

They are equal .

QPages 165-170: Q42: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q42 Solution: A metal sphere of radius R is charged to a potential V. 1. Find …

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