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Physics Question 40 – JEE-MAIN 2026

Two point charges 8μC and 2μC 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 _______.

Gauss's Law states that the total electric flux through any closed surface is proportional to the net electric charge enclosed within that surface.

Step 1: Calculate flux through the first sphere✦ Active

The first sphere has a radius of R1=3 cm and is centered at the origin. The charges are q1=8μC at x=2 cm and q2=2μC at x=4 cm.

Charge q1 is at x=2 cm, which is less than R1=3 cm, so q1 is inside the sphere. Charge q2 is at x=4 cm, which is greater than R1=3 cm, so q2 is outside the sphere.

According to Gauss's Law, the total charge enclosed by the first sphere is Qenclosed,1=q1=8μC. The electric flux through the first sphere is:

Φ1=Qenclosed,1ϵ0=8μCϵ0
Step 2: Calculate flux through the second sphere○ Expand

The second sphere has a radius of R2=5 cm and is centered at the origin.

Charge q1 is at x=2 cm, which is less than R2=5 cm, so q1 is inside the sphere. Charge q2 is at x=4 cm, which is less than R2=5 cm, so q2 is also inside the sphere.

The total charge enclosed by the second sphere is Qenclosed,2=q1+q2=8μC+(2μC)=6μC. The electric flux through the second sphere is:

Φ2=Qenclosed,2ϵ0=6μCϵ0
Step 3: Determine the ratio of electric fluxes○ Expand

The ratio of electric flux through the two spheres is:

Φ1Φ2=8μCϵ06μCϵ0=86=43

The ratio is 4:3.

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