JEE Main 2014 · 12 Apr · Shift 1 · Physics Q17

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PhysicsSingle correct+4 / -1Question 17 of 90

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation :
$K(x)=K_o+\lambda x\quad(\lambda=\text{ a constant})$
The capacitance C, of this capacitor, would be related to its 'vacuum' capacitance $C_o$ as per the relation :

C=\dfrac{\lambda d}{\ln(1+K_o\lambda d)}C_o

C=\dfrac{\lambda}{d\cdot\ln(1+K_o\lambda d)}C_o

C=\dfrac{\lambda d}{\ln(1+\lambda d/K_o)}C_o

C=\dfrac{\lambda}{d\cdot\ln(1+K_o/\lambda d)}C_o

Answer and worked solution

Official answer: C

Thin slabs in series:
$\frac1C=\int_0^d\frac{dx}{\epsilon_oA(K_o+\lambda x)}=\frac1{\epsilon_oA\lambda}\ln\frac{K_o+\lambda d}{K_o}.$
With $C_o=\frac{\epsilon_oA}d$ :
$C=\frac{\lambda d}{\ln(1+\lambda d/K_o)}C_o.$

Topics:Dielectrics

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 17