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A ring of radius R is with a uniformly distributed charge on it. A charge q is now placed at the centre of the ring. Find the increment in tension in the ring. |
Answer» Solution :  Consider an ELEMENT of the ring. Its enlarged view is as SHOWN. For equilibrium of this segment, we can write. `F = 2 DeltaT sin ((d theta)/(2))` Here F is the repulsive force between q and elemental CHARGE `dQ [ :. dQ = Q/(2 pi R) (Rd theta)]` The electric outward force on element is `F = (1)/(4 pi epsilon_0) (q dQ)/(R^2)` From the above three EQUATIONS, we can write, `1/(4 pi epsilon_0) q/(R^2) (QRd theta)/(2 pi R) ~~ 2 Delta T ((d theta)/(2)) ( :. sin alpha~~ alpha` for SMALL angle) `Delta T = (Qq)/(8 pi^2 epsilon_0 R^2)` |
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