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A ring of radius R is with a uniformly distributed charge Q 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 Delta T sin ((d theta)/(2))` Here F is the repulsive force between q and elemental charge `DQ [ because dQ = (Q)/(2pi R) (R d 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)) (QR d theta)/(2pi R) ~~ 2 Delta T ((d theta)/(2)) ( because sin ALPHA ~~ alpha " for small ANGLE")` `Delta T= (Qq)/(8pi^(2) in_(0)R^(2))` |
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