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← JEE Main 2026 · 2 Apr · Shift 1
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JEE Main 2026 · 2 Apr · Shift 1, Physics question 34

Two charged conducting spheres S1S_1 and S2S_2 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 S1S_1 and S2S_2 spheres are ES1E_{S1} and ES2E_{S2} respectively. The value of ES1ES2\dfrac{E_{S1}}{E_{S2}} is ________.

Choose one option
Answer and worked solution

Official answer: D

Same potential. V=kQrV=\dfrac{kQ}r is equal for both spheres, and the surface field is E=kQr2=VrE=\dfrac{kQ}{r^2}=\dfrac Vr.

Ratio. ES1ES2=r2r1=188=94\dfrac{E_{S1}}{E_{S2}}=\dfrac{r_2}{r_1}=\dfrac{18}8=\dfrac94.

(D)