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

Concept: According to Gauss’s Law, the total electric flux through a closed surface is: \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \] Thus, electric flux depends only on the net charge enclosed inside the surface.

Step 1: Identify the charges enclosed by the sphere of radius \(3\,\text{cm}\). Charges are located at: \[ +8\,\mu C \text{ at } x=2\,\text{cm} \] \[ -2\,\mu C \text{ at } x=4\,\text{cm} \] For sphere of radius \(3\,\text{cm}\):

• \(+8\,\mu C\) lies inside
• \(-2\,\mu C\) lies outside

Hence enclosed charge: \[ q_1 = 8\,\mu C \] Therefore, flux through first sphere: \[ \Phi_1 = \frac{8}{\varepsilon_0} \]

Step 2: Identify the charges enclosed by the sphere of radius \(5\,\text{cm}\). For sphere of radius \(5\,\text{cm}\):

• Both charges lie inside

Net enclosed charge: \[ q_2 = 8 + (-2) \] \[ q_2 = 6\,\mu C \] Therefore, flux through second sphere: \[ \Phi_2 = \frac{6}{\varepsilon_0} \]

Step 3: Find the ratio of electric fluxes. \[ \Phi_1 : \Phi_2 = \frac{8}{\varepsilon_0} : \frac{6}{\varepsilon_0} \] \[ = 8:6 \] \[ = 4:3 \] Therefore, \[ \boxed{4:3} \]