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A thin spherical shell radius of r has a charge Q uniformly distributed on it. At the centre of the shell, a negative point charge -q is placed. If the shell is cut into two identical hemispheres, still equilibrium is maintained. Then find the relation between Q and q? |
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Answer» Solution :Here the outward electric pressure at every point on the shell due to its own charge is `P_(1) = (sigma^(2))/(2in_(0)) =(1)/(2 in_(0)) ((Q)/(4pi r^(2)))^(2), P_(1) = (Q^(2))/(32pi^(2) in_(0)r^(4))` Due to -q, the electric field on the surface of the shell is `E= (1)/(4pi in_(0)) (q)/(r^(2))` This electric field pulls every point of the shell in INWARD direction. The inward pressure on the surface of the shell due to the negative charge is `P_(2)= sigma E` `= ((Q)/(4pi r^(2))) ((1)/(4pi in_(0))(q)/(r^(2)))= (Qq)/(16pi^(2) in_(0)r^(4))` For equilibrium of the HEMISPHERICAL shells `P_(2) ge P_(1)` or `(Qq)/(16pi^(2) in_(0)r^(4)) ge (Q^(2))/(32pi^(2) in_(0)r^(4)) q ge (Q)/(2)` |
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