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Class 12 › Physics › Electric Charge and Field
CBSE2024Class Class 12 · Physics2 Marks · Short✅ Verified
A conducting wire connects two charged conducting spheres of radii r₁ and r₂ such that they attain equilibrium with respect to each other. The distance of separation between the two spheres is very large as compared to either of their radii. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres of radii r₁ and r₂.
✅ Answer & Solution
<b>Concept:</b> When two conductors are connected by a wire, they reach the <b>same potential</b>.<br><br><b>Step 1: Equate potentials</b><br>V<sub>1</sub> = V<sub>2</sub><br>kq<sub>1</sub>/r<sub>1</sub> = kq<sub>2</sub>/r<sub>2</sub><br><br>Therefore: q<sub>1</sub>/q<sub>2</sub> = r<sub>1</sub>/r<sub>2</sub><br><br><b>Step 2: Electric field at surface</b><br>E = kq/r²<br><br><b>Step 3: Take ratio</b><br>E<sub>1</sub>/E<sub>2</sub> = (q<sub>1</sub>/q<sub>2</sub>) × (r<sub>2</sub>/r<sub>1</sub>)²<br>E<sub>1</sub>/E<sub>2</sub> = (r<sub>1</sub>/r<sub>2</sub>) × (r<sub>2</sub>/r<sub>1</sub>)²<br>E<sub>1</sub>/E<sub>2</sub> = <b>r<sub>2</sub>/r<sub>1</sub></b><br><br><b>Answer:</b> E<sub>1</sub> : E<sub>2</sub> = <b>r<sub>2</sub> : r<sub>1</sub></b><br><br><b>Conclusion:</b> Smaller sphere has stronger surface field.
