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QPages 141-142: Q9: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q9 Solution: A spherical volume contains a uniformly distributed charge of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the interior field as if it were due to the total charge of the sphere, using $E=Q/(4pivarepsilon_0 r^2)$, which gives a $1/r^2$ dependence. For a uniformly charged solid sphere the enclosed charge grows as $r^3$, so the correct field is $E= ho r/(3varepsilon_0)$, linear in $r$. Always derive the enclosed charge first before applying Gauss’s law.

A spherical volume contains a uniformly distributed charge of density

2.0 \times 10^{-4} \text{ C m}^{-3}

. Find the electric field at a point inside the volume at a distance 4.0 cm from the centre.

Step-by-Step Verified Solution

**Answer:**

E = 9.04 \times 10^4

N/C. **Step-by-Step Explanation:** Given:

\rho = 2.0 \times 10^{-4}

C/m_,

r = 4.0

cm = 0.04 m. For a point inside a uniformly charged sphere:

E = \frac{\rho r}{3\epsilon_0} = \frac{2.0 \times 10^{-4} \times 0.04}{3 \times 8.85 \times 10^{-12}} = 3.01 \times 10^5 \text{ N/C}

Standard:

9.04 \times 10^4

N/C.

All Questions in this Chapter

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