Question

Two point charges \(8\,\mu C\) and \(-2\,\mu C\) are located at \(x = 2\,\text{cm}\) and \(x = 4\,\text{cm}\), respectively on the \(x\)-axis. The ratio of electric flux due to these charges through two spheres of radii \(3\,\text{cm}\) and \(5\,\text{cm}\) with their centers at the origin is ________.

• A. \(4:1\)
• B. \(3:4\)
• C. \(4:3\)
• D. \(4:5\)

Hint:

Remember:

• Electric flux depends only on enclosed charge
• According to Gauss’s law: \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \]
• Charges outside the Gaussian surface do not contribute to net flux

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Electric flux through a closed surface is a measure of the total number of electric field lines passing through that surface.
Gauss's Law provides a powerful way to calculate this flux without performing complex surface integrals.
It states that the net flux (\(\Phi\)) through any closed Gaussian surface is equal to the net charge enclosed by the surface (\(q_{enclosed}\)) divided by the permittivity of free space (\(\epsilon_0\)).
Crucially, charges located outside the surface contribute zero to the net flux, as any field lines entering the volume from an external charge must also leave it.
The shape or size of the surface does not matter, as long as it is closed; only the magnitude and sign of the internal charges determine the flux.
Key Formula or Approach:
1. Gauss's Law: \[ \Phi = \frac{q_{enclosed}}{\epsilon_0} \]
2. Identification of enclosed charges based on radius \(r\).
Step 2: Detailed Explanation:
We have two point charges: \(Q_1 = +8\,\mu\)C at \(x = 2\text{ cm}\) and \(Q_2 = -2\,\mu\)C at \(x = 4\text{ cm}\).
The spheres are centered at the origin \((0,0)\).
Case 1: Sphere of radius \(R_1 = 3\text{ cm}\)
The charge at \(x = 2\text{ cm}\) is within the radius (\(2<3\)), so it is enclosed.
The charge at \(x = 4\text{ cm}\) is outside the radius (\(4>3\)), so it is not enclosed.
Net enclosed charge \(q_1 = +8\,\mu\)C.
Flux \(\Phi_1 = \frac{8\,\mu\text{C}}{\epsilon_0}\).
Case 2: Sphere of radius \(R_2 = 5\text{ cm}\)
The charge at \(x = 2\text{ cm}\) is within the radius (\(2<5\)), so it is enclosed.
The charge at \(x = 4\text{ cm}\) is also within the radius (\(4<5\)), so it is enclosed.
Net enclosed charge \(q_2 = +8\,\mu\text{C} + (-2\,\mu\text{C}) = +6\,\mu\)C.
Flux \(\Phi_2 = \frac{6\,\mu\text{C}}{\epsilon_0}\).
Ratio Calculation:
The ratio of fluxes is the ratio of the enclosed charges:
\[ \text{Ratio} = \frac{\Phi_1}{\Phi_2} = \frac{8}{6} \]
Simplifying the fraction by dividing both the numerator and denominator by \(2\):
\[ \text{Ratio} = \frac{4}{3} \]
Step 3: Final Answer:
The ratio of the electric fluxes is \(4 : 3\).