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A ringof radius R is with uniformly distributed charge Q on it. A charge q is now placed at the centreof the ring. Find the increment in tension in the ring. |
Answer» SOLUTION : CONSIDER an element. Its enlarged view is as shown. For equilibrium of this segment, we can write. `F= 2DeltaT sin ((d THETA)/(2))` Here F is the repulsive FORCE between q and elementalcharge dQ `dQ=(Q)/( 2pi R) (Rd theta)` The electric outward force on element is `F=(1)/(4pi in_(0)) (qdQ)/( R^(2))` From the above three equations, we can write `(1)/(4pi in_(0)) (q)/(R^(2))(QRd theta)/(2piR) ~~ 2DeltaT((d theta)/(2))` (`:. sin alpha = alpha` for small ANGLE) `Delta = (Qq)/( 8pi^(2) in_(0) R^(2))` |
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