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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Gauss's Law

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Gauss's Law

Determine the flux of the electric field through a spherical...

Determine the flux of the electric field through a spherical surface of radius RR caused by a charge of 10−7 C10^{-7} \text{ C} located at its centre and a second identical charge positioned at a distance of 2R2R from the centre, as shown in the figure.
Determine the flux of the electric field through a spherical surface of radius R caused by a charge of 10^ -7 C located

Options

  1. A1.1×104 N m2 C−11.1 \times 10^4 \text{ N m}^2 \text{ C}^{-1}
  2. B00
  3. C0.55×104 N m2 C−10.55 \times 10^4 \text{ N m}^2 \text{ C}^{-1}
  4. D2.2×104 N m2 C−12.2 \times 10^4 \text{ N m}^2 \text{ C}^{-1}

Correct answer

A. 1.1×104 N m2 C−11.1 \times 10^4 \text{ N m}^2 \text{ C}^{-1}

Step-by-step solution

According to Gauss's Law, the total electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space (ε0\varepsilon_0).

Φ=qenclosedε0\Phi = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}

The spherical surface has a radius RR. The charge located at the centre is inside the sphere, while the identical charge positioned at a distance 2R2R from the centre is outside the sphere.

Therefore, the net charge enclosed by the spherical surface is only the charge at the centre.

qenclosed=10−7 Cq_{\text{enclosed}} = 10^{-7} \text{ C}

The electric flux through the spherical surface is:

Φ=10−78.85×10−12\Phi = \dfrac{10^{-7}}{8.85 \times 10^{-12}}

Φ≈1.129×104 N m2 C−1\Phi \approx 1.129 \times 10^4 \text{ N m}^2 \text{ C}^{-1}

Rounding to appropriate significant figures, we get:

Φ≈1.1×104 N m2 C−1\Phi \approx 1.1 \times 10^4 \text{ N m}^2 \text{ C}^{-1}
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