JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2026
- Medium
- MCQ
Two point charges \(8 \, \mu C\) and \(-2 \, \mu C\) are located at \(x = 2\) cm and \(x = 4\) cm, respectively on the \(x\)-axis. The ratio of electric flux due to these charges through two spheres of radii \(3\) cm and \(5\) cm with their centers at the origin is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4 : 3\)
Step-by-step Solution
Detailed explanation
According to Gauss's law, the electric flux through a closed surface is given by \(\Phi = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}\).
For the first sphere of radius \(r_1 = 3\) cm centered at the origin, only the charge at \(x = 2\) cm is enclosed.
\(q_{\text{enclosed}, 1} = 8 \, \mu\text{C}\)
\(\Phi_1 = \dfrac{8 \, \mu\text{C}}{\varepsilon_0}\)
For the second sphere of radius \(r_2 = 5\) cm centered at the origin, both charges at \(x = 2\) cm and \(x = 4\) cm are enclosed.
\(q_{\text{enclosed}, 2} = 8 \, \mu\text{C} - 2 \, \mu\text{C} = 6 \, \mu\text{C}\)
\(\Phi_2 = \dfrac{6 \, \mu\text{C}}{\varepsilon_0}\)
The ratio of the electric flux is:
\(\dfrac{\Phi_1}{\Phi_2} = \dfrac{8}{6} = \dfrac{4}{3}\)
Answer: \(4 : 3\)
For the first sphere of radius \(r_1 = 3\) cm centered at the origin, only the charge at \(x = 2\) cm is enclosed.
\(q_{\text{enclosed}, 1} = 8 \, \mu\text{C}\)
\(\Phi_1 = \dfrac{8 \, \mu\text{C}}{\varepsilon_0}\)
For the second sphere of radius \(r_2 = 5\) cm centered at the origin, both charges at \(x = 2\) cm and \(x = 4\) cm are enclosed.
\(q_{\text{enclosed}, 2} = 8 \, \mu\text{C} - 2 \, \mu\text{C} = 6 \, \mu\text{C}\)
\(\Phi_2 = \dfrac{6 \, \mu\text{C}}{\varepsilon_0}\)
The ratio of the electric flux is:
\(\dfrac{\Phi_1}{\Phi_2} = \dfrac{8}{6} = \dfrac{4}{3}\)
Answer: \(4 : 3\)
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