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Two identical metal plates are positively charged to Q_1" and " Q_2(Q_2 lt Q_1). If they are brough near each other to form a parallel plate capacitor of capacitance C, then what will be the potential differences between the plates? |
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Answer» SOLUTION :Let the area of the metal plates be A and intensity of the electric fields at any point between the plates due to the FIRST and second metal plates be `E_1 " and "E_2` RESPECTIVELY. Electric field at any between the plates due to first plate, `E_1 = (Q_1)/(2A in_0)` Electric field at any point between the plates due to second plate, `E_2 = (Q_2)/(2A in_0)` So net electric field, `E = E_1 - E_2 = ((Q_1 - Q_2)/(2A in_0))""[" where "in_0 = " permittivity of free space "]` Again capacitance of parallel plate capacitor, `C= (in_0 A)/(d) [ d= " distance between the two plates "]` We know, POTENTIAL difference = net electric field `XX` distance `:. V = (E_1 - E_2)d = ((Q_1 -Q_2)/(2A in_0))d = ((Q_1 - Q_2)/(2))(d)/(A in_0)` Hence potential difference, `V = (Q_1 - Q_2)/(2C) [ :. C= (in_0 A)/(d)]`. |
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