SATHEE: Electrostatics Question 804

Electrostatics Question 804

Question: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$

Options:

A) 72 pF

B) 81 pF

C) 84 pF

D) 96 PF

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Capacity of parallel plate capacitor

$C = \frac{ε_{r} ε_{0} A}{d} (For air ε_{r} = i)$

So, $\frac{ε_{0} A}{d} = 8 \times 10^{- 12}$

If $d \to \frac{d}{2}$

and $ε_{r} \to 6$

then new capacitance $C^{'} = 6 \times \frac{ε_{0} A}{d / 2} = 12 \frac{ε_{0} A}{d} = 12 \times 8 p F = 96 p F$

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Electrostatics Question 804

Question: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. (εr=6)

Options:

Answer:

Solution:

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Question: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$