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In the last illustration, suppose that the battery is kept connected while the plates are pulled apart. What are the answers to the parts (a), (b), (c) and (d) in that case ? |
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Answer» Solution : If the battery is kept connected, the potential difference ACROSS the capacitor plates always remains equal to the emf of battery and HENCE is constant. ` V =V_0 = 50 V ` ` C =( in _0 A)/(d) =( in _0 A)/( 2d) =(C_0)/(2) 1.77 xx 10^(-1) MUF ` ` Q =CV =(C_0 V_0)/(2) =(Q_0)/(2)=(Q_0)/(2) = 8.85 xx 10 ^(-4) mu C ` ` U= (1)/(2) CV^(2) =(1)/(2) ((C_0)/( 2)) V_0^(2) =(U_0)/(2) =2.21xx 10 ^(-5) J` `E =(V)/(d) =(V_0)/(2d_0)=(E_0)/(2) = 2500 V//m ` |
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