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A parallel plate air capacitor is made using two square plates each of side 0.2 m, spaced 1 cm apart. It is connected to a 50V battery. (a) What is the capacitance ? (b) What is the charge on each plate ? (c) What is the energy stored in the capacitor ? (d) What is the electric field between the plates ? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 2 cm, what are the answers to the above parts ? |
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Answer» Solution :` C_0 =(in _0A)/(d) =( (8.85xx 10 ^(-12) xx 0.2 xx 0.2 )/( 0.01) = 3.54xx 10^(-5) mu F ` (b)`Q_0 =C_0 V_0 = 3.54 xx 10^(-5)xx 50= 1.77 xx 10^(-3)mu C ` (c) ` U_0=(1)/(2)C_0 V_0^(2) =(1)/(2)xx ( 3.54xx 10^(-11) ) ( 50)^(2)= 4.42 xx 10 ^(-4) J` (d) ` E_0 =(V_0)/(d) =( 50) /( 0.01) = 5000 V //m ` If the battery is disconnected the charge on the capacitor plates remains constant while the potential DIFFERENCE between the plates can CHANGE. `C= (in _0A) /(d)=(C_0)/( 2) =1.77xx 10^(-5) mu F . ` ` Q= Q_0 1.77 xx 10^(3) mu C ` `V= (Q)/(C) =( Q_0)/(C_0//2) =2V_0` ` U= (1)/(2) CV^(2) =C_0 V_0^(1) =8.84 xx 10 ^(-3)J` ` E= (V)/(d) =(2V_0)/( 2d_0) =E_0 =5000 V//m` |
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