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Physics Question 34 – JEE-MAIN 2026

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Two charged conducting spheres S1 and S2 of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on S1 and S2 spheres are ES1 and ES2 respectively. The value of ES1ES2 is _______.

When conducting spheres are connected by a wire, charge redistributes until their electric potentials become equal.

Step 1: Apply Electrostatic Equilibrium Condition✦ Active

When two conducting spheres are connected by a wire, they reach electrostatic equilibrium, meaning their electric potentials become equal. Let this common potential be V.

V1=V2=V
Step 2: Express Electric Field in terms of Potential and Radius○ Expand

The electric field on the surface of a conducting sphere of radius R and potential V is given by E=VR.

For sphere S1 with radius R1=8 cm: ES1=VR1

For sphere S2 with radius R2=18 cm: ES2=VR2

Step 3: Calculate the Ratio of Electric Fields○ Expand

The ratio of the electric fields is:

ES1ES2=V/R1V/R2=R2R1

Substitute the given radii R1=8 cm and R2=18 cm:

ES1ES2=188=94
💡 Teacher's Secret Hint

Ensure units are consistent, though in a ratio, they cancel out.

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