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QPages 141-142: Q7: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q7 Solution: Find the flux of the electric field through a spherical surface …

Find the flux of the electric field through a spherical surface of radius R due to a charge of 10−7 C10^{-7} \text{ C} at the centre and another equal charge at a point 2R away from the centre in the following figure.

Step-by-Step Verified Solution

**Answer:** Flux = 10−7ϵ0=1.13×104\frac{10^{-7}}{\epsilon_0} = 1.13 \times 10^4 N m_/C. **Step-by-Step Explanation:** The spherical surface of radius R encloses only the charge at the centre (10−710^{-7} C). The other charge at distance 2R is outside the surface and contributes zero net flux.
Φ=Qenclosedϵ0=10−78.85×10−12=1.13×104 N m2/C\Phi = \frac{Q_{enclosed}}{\epsilon_0} = \frac{10^{-7}}{8.85 \times 10^{-12}} = 1.13 \times 10^4 \text{ N m}^2/\text{C}

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