SATHEE: Electrostatics Question 809

Electrostatics Question 809

Question: The energy required to change a parallel plate condenser of plate separation d and plated area of cross-section A such that the uniform electric field between the plates is E, is

Options:

A) $\in_{0} E^{2} A d$

B) $\frac{1}{2} \in_{0} E^{2} A d$

C) $\frac{1}{2} \in_{0} E^{2} / A d$

D) $ε_{0} E^{2} / A d$

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Answer:

Correct Answer: A

Solution:

[a] Energy required to charge the capacitor is $W = U = Q V$

$\Rightarrow U = C V^{2} = \frac{ε_{0} A}{d} y^{2} = \frac{ε_{0} A d}{d^{2}}$

$V^{2} = ε_{0} E^{2} A d [∵ E = \frac{V}{d}]$

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Electrostatics Question 809

Question: The energy required to change a parallel plate condenser of plate separation d and plated area of cross-section A such that the uniform electric field between the plates is E, is

Options:

Answer:

Solution:

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