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A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R ( R gt gt r) , such that their surface charge densities are equal . Derive the expression for the potential at the common centre. |
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Answer» Solution :As shown let charges on inner and outer spheres be `Q_(1)` and `Q_(2)` respectively , where `Q_(1) + Q_(2) = Q` Let COMMON surface CHARGE density be `sigma` , then ` Q_(1) = 4 pi r^(2) sigma` and `Q_(2) = 4 pi R^(2) sigma` `implies Q = Q_(1) + Q_(2) = [4 pi r^(2) + 4 pi R^(2)] sigma = 4 pi sigma [r^2 + R^2] implies sigma = (Q)/(4pi [r^(2) + R^(2)])` THUS , `Q_1 = (4pi r^(2) * Q)/(4pi [r^(2) + R^(2)]) = (Qr^(2))/((r^(2) + R^(2)))` and `Q_(2) = (4pi R^(2) * Q)/(4pi [r^(2) + R^(2)]) = (Q R^(2))/((r^(2) + R^(2)))` Now potential at centre point O due to charge`Q_(1)` on the HOLLOW sphere of radius r is , `V_(1) = (Q_(1))/(4pi in_(0) r) = (Q r^(2))/(4 pi in_(0) r ( r^(2) + R^(2))) = (Qr)/(4pi in_(0) (r^(2) + R^(2)))` and potential at centre point O due to charge `Q_2` on the hollow sphere of radius R is , `V_(2) = (Q_(2))/(4pi in_(0) R) = (Q R^(2))/(4 pi in_(0) R (r^(2) + R^(2))) = (Q . R)/(4 pi in_(0) (r^(2) + R^(2)))` `therefore` Total potential at centrepoint O `V = V_(1) + V_(2) = (Q r)/(4 pi in_(0) (r^(2) + R^(2))) + (Q. R)/(4 pi in_(0) (r^(2) + R^(2))) = (Q)/(4pi in_(0)) * ((r + R))/((r^(2) + R^(2)))` 
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