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Question
Two parallel-plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates, while Y contains a dielectric medium of εr = 4.
(i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.
(ii) Calculate the potential difference between the plates X and Y.
(iii) Estimate the ratio of electrostatic energies stored in X and Y.
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Solution
(i) We know that the capacitance of a parallel-plate capacitor is given by
`C=(varepsilon_0varepsilon_rA)/d`
`C_Y=(varepsilon_0 4A)/d`
`C_X=(varepsilon_0 A)/d`
Thus,
`=>C_Y=4C_X`
`=>C_(eq)=(C_xC_Y)/(C_x+C_Y) = 4muF`
`=>(C_X(4C_X))/(5C_X)=4muF`
`=>C_X=5muF`
`=>C_Y=20muF`
(ii) The potential difference between plates X and Y can be calculated as follows:
Q=CV
`=>C_XV_X=C_YV_Y`
`=>V_X/V_Y=C_X/C_Y=4`
`=>V_X=4V_Y`
Also, VY+VX=15
⇒VY=3 V
⇒VX=12 V
(iii) The ratio of electrostatic energies can be calculated as follows:
`E=(Q^2)/(2C)`
`=>E_X/E_Y=C_Y/C_X=4`
`=>E_X/E_Y=4/1`
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