Verified Solution • 2025–26 Session moderate

QJEEM24Ph2754: JEE Main 2024 Physics - Class 12 JEEM24Ph2754 Solution: Two charges of and are situated at the points and respectively. …

Two charges of 5Q5 Q and −2Q-2 Q are situated at the points (3a,0)(3 a, 0) and (−5a,0)(-5 a, 0) respectively. The electric flux through a sphere of radius ' 4a4 a ' having center at origin is :

Multiple Choice Options

  • A 2Qε0\frac{2 Q}{\varepsilon_0}
  • B 7Qε0\frac{7 Q}{\varepsilon_0}
  • C 3Qε0\frac{3 Q}{\varepsilon_0}
  • D 5Qε0\frac{5 Q}{\varepsilon_0}

Step-by-Step Verified Solution

The electric flux through any closed surface is given by Gauss's law, which can be stated as:

Φ=Qencε0\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}

where:

  • Φ\Phi = electric flux
  • QencQ_{\text{enc}} = total charge enclosed by the surface
  • ε0\varepsilon_0 = permittivity of free space (electric constant)

In this scenario, we have a sphere of radius 4a4a with its center at the origin and two charges, 5Q5Q at point (3a, 0) and −2Q-2Q at point (-5a, 0). Since the sphere's radius is 4a4a , the charge 5Q5Q , which is located at (3a, 0), lies inside the sphere, whereas the charge −2Q-2Q , located at (-5a, 0), lies outside the sphere. Gauss's law only considers charges that are enclosed within the surface.

Thus, the charge −2Q-2Q has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge 5Q5Q contributes to the electric flux inside the sphere.

The electric flux through the sphere is therefore given simply by the enclosed charge:

Φ=5Qε0\Phi = \frac{5Q}{\varepsilon_0}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring