SATHEE: Electrostatics Question 639

Electrostatics Question 639

Question: 4. A capacitor is filled with dielectrics. The dii resultant capacitance is

Options:

A) $\frac{2 ε_{0} A}{d} [\frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}}]$

B) $\frac{ε_{0} A}{d} [\frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}}]$

C) $\frac{2 ε_{0} A}{d} [K_{1} + K_{2} + K_{3}]$

D) None of these

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $C_{1} = \frac{K_{1} ε_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{K_{1} ε_{0} A}{d}$

$C_{2} = \frac{K_{2} ε_{0} (\frac{A}{2})}{(\frac{d}{2})} = \frac{K_{2} ε_{0} A}{d}$

and $C_{3} = \frac{K_{3} ε_{0} A}{2 d}$

Now, $C_{e q} = C_{3} + \frac{C_{1} C_{2}}{C_{1} + C_{2}} = (\frac{K_{3}}{2} + \frac{K_{1} K_{2}}{K_{1} + K_{2}}) . \frac{ε_{0} A}{d}$

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