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A ring of radius R is having charge Q uniformly distributed. A point charge Q_(0) is placed at the centre of the ring. Find tension developed in the ring. If radius of cross - section of the ring is a and Young's modulus of wire is Y, find increase in the radius of ring. |
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Answer» Solution :To find tension//compression in the ring, find the force on small length of angle ` d theta` and remove `d theta` from the expression (refer to Circular Motion , Example 33 , page 351 Vol. 1). Charge on small length of angle `d theta` `Q' = (Q)/( 2 pi R) DL = (Q)/( 2 pi R) R d theta = (Q d theta)/(2 pi)` Electric force on `Q'` , `F = (1)/(4 pi in_(0)) (Q' Q_(0))/(R^(2)) = (1)/( 4 pi in_(0)) . (Q Q_(0))/(2 pi R^(2)) d theta` Tension in the string `T = (Q Q_(0))/( 8 pi^(2) in_(0) R^(2))` Let increase in the radius of ring is `DELTA R`. Longitudinal strain `in = ( 2pi (R + Delta R) - 2 pi R)/( 2 pi R) = (Delta R)/(R)` Longitudinal stress `sigma = (T)/(A) = (T)/(pi d^(2))` where `a`: radius of cross - section of ring. `Y = (sigma)/(in) = (T//pi a^(2))/(Delta R//R)` `Delta R = (T R)/(pi a^(2) Y) = (Q Q_(0))/(8 pi^(2) in_(0) R^(2)) . (R )/(pi a^(2) Y)` `= (Q Q_(0))/(8 pi^(3) in_(0) R Y a^(2))`  
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