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Electrostatics

JEE Main level, single correct, medium

279 of 311

Question 279 of 311

${C}_{1}$ and ${C}_{2}$ are two hollow concentric cubes enclosing charges $2Q$ and $3Q$ respectively as shown in figure. The ratio of electric flux passing through ${C}_{1}$ and ${C}_{2}$ is:

Show answer and solution

Answer: (A) $2:5$

Gauss's law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. Therefore,

{\phi }_{\text{smaller cube}}=\dfrac{2Q}{{\epsilon }_{0}}

Since the smaller cube is inside the bigger cube, so total charge contained by the bigger cube will be

2Q+3Q=5Q

{\phi }_{\text{bigger cube}}=\dfrac{5Q}{{\epsilon }_{0}}

Therefore, required ratio will be,

\dfrac{{\phi }_{\text{smaller cube}}}{{\phi }_{\text{bigger cube}}}=\dfrac{2}{5}

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electrostatics (Electric Flux and Gauss Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

C1{C}_{1}C1​ and C2{C}_{2}C2​ are two hollow concentric cubes enclosing charges 2Q2Q2Q and 3Q3Q3Q respectively as shown in figure. The ratio of electric flux passing through C1{C}_{1}C1​ and C2{C}_{2}C2​ is:

JEE Main question paper PDF

More Physics practice

${C}_{1}$ and ${C}_{2}$ are two hollow concentric cubes enclosing charges $2Q$ and $3Q$ respectively as shown in figure. The ratio of electric flux passing through ${C}_{1}$ and ${C}_{2}$ is: