Question

A parallel plate capacitor has sqare plates of side 5cm and seperated by a distance of 1mm the capacity of capacitor is

Answer 1

$\huge{\underline{\underline{\boxed{\sf{\green{Answer :- }}}}}}$

$Given \: : \:   d=2  \:  mm   \:    \\   \\     E \: = \: 3.0 \: × \: 104 \:  V/cm \: = \: 3.0 \: × \: 106  \:  V/m  \\  \\       l=5.0cm=0.05 m</p><p></p><p>Area \: of \: each \: plat \: \\ \\      A \: = \: \: = \: (0.05)2 \: = \: 25×10−4  m2 \\ </p><p> </p><p>Capacitance \: of \: the \: capacitor   \: \\ \\       C=dAϵo</p><p></p><p></p><p>Potential \: difference \: between \: the \: plates         V \: = \: E d \\ \\</p><p></p><p>∴   Charge \: on \: each \: plate \:   \\ \\    Q=CV=AϵoE \\ \\</p><p></p><p>⟹   \:  Q=25×10−4×8.85×10−12×3.0×106=6.6×10−8  C$

Answer 2

Answer:

\huge{\underline{\underline{\boxed{\sf{\green{Answer :- }}}}}}

Answer:−

\begin{gathered} Given \: : \: d=2 \: mm \: \\ \\ E \: = \: 3.0 \: × \: 104 \: V/cm \: = \: 3.0 \: × \: 106 \: V/m \\ \\ l=5.0cm=0.05 m < /p > < p > < /p > < p > Area \: of \: each \: plat \: \\ \\ A \: = \: \: = \: (0.05)2 \: = \: 25×10−4 m2 \\ < /p > < p > < /p > < p > Capacitance \: of \: the \: capacitor \: \\ \\ C=dAϵo < /p > < p > < /p > < p > < /p > < p > Potential \: difference \: between \: the \: plates V \: = \: E d \\ \\ < /p > < p > < /p > < p > ∴ Charge \: on \: each \: plate \: \\ \\ Q=CV=AϵoE \\ \\ < /p > < p > < /p > < p > ⟹ \: Q=25×10−4×8.85×10−12×3.0×106=6.6×10−8 C \\ \end{gathered}

Given: d=2 mm

E=3.0×104 V/cm=3.0×106 V/m

l=5.0cm=0.05 m</p><p></p><p>Areaofeachplat

A==(0.05)2=25×10−4 m2

</p><p></p><p>Capacitanceofthecapacitor

C=dAϵo</p><p></p><p></p><p>Potentialdifferencebetweentheplates V=Ed

</p><p></p><p>∴ Chargeoneachplate

Q=CV=AϵoE

</p><p></p><p>⟹ Q=25×10−4×8.85×10−12×3.0×106=6.6×10−8 C