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

Question

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E , is (A)  (1/2) ε0 E2 /Ad  (B)  ε0 E2/Ad  (C)  ε0 E2Ad  (D)  (1/2) ε0 E2Ad

Tardigrade Admin · Accepted Answer

Energy given by the cell E=C V2 Here, C= capacitance of condenser =(A ε0/d) V= potential difference across the plates =E d Therefore, E=((A ε0/d))(E d)2 =A ε0 E2d

Q. The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$ , is

$\frac{1}{2} ε_{0} E^{2} / A d$

$ε_{0} E^{2} / A d$

$ε_{0} E^{2} A d$

$\frac{1}{2} ε_{0} E^{2} A d$

Energy given by the cell
$E = C V^{2}$
Here, $C =$ capacitance of condenser
$= \frac{A ε _{0}}{d}$
$V =$ potential difference across the plates $= E d$
Therefore, $E = (\frac{A ε _{0}}{d}) (E d)^{2}$
$= A ε_{0} E^{2} d$

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

21​ε0​E2/Ad

ε0​E2/Ad

ε0​E2Ad

21​ε0​E2Ad