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Two isolated metallic solid spheres fo radii R and 2R are charged in such a way that both of these have same charge density sigma. The spheres are placed far away from each other and are connected by a thin conducting wire. Find the new charge density on the bigger sphere. |
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Answer» Solution :LET `sigma` be the CHARGE density of the two sphere. So, charge of the fisrt sphere `=q_1= 4pi R^2 sigma` and charge of the second sphere `= q_2= 4 pi (2R)^2 sigma = 16 pi R^2 sigma` When they are connected with a wire, let `q_1'" and "q_2'` be the new charges. Then we may write `q_1'+q_2'= q_1+q_2"""................"(1)` Since the two spheres are at the same potential `(1)/(4 pi in_0R)(q_1')/(R )= (1)/(4pi in_0)(q_2')/(2R)" or "q_1'= (q_2')/(2)` In the equation (1), by substituting `q_1'`, we have `(q_2')/(2)+q_2'= q_1+q_2` or `q_2'= (2)/(3)(q_1+q_2)" or "q_2'=(2)/(3)(4pi R^2 sigma +16 pi R^2 sigma)` `:. ""q_2' = (40 piR^2 sigma)/(3)` Therefore new charge density of the bigger sphere, `sigma' = (q_2')/(4 pi(2R)^2)= (40piR^2 sigma)/(3xx16xxpi R^2)= (5)/(6) sigma`. |
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