Question 501: Electrostatics & Capacitance

Question

Charge on a parallel plate capacitor of capacity C is Q , the electric field intensity between its two plates separated by a distance of $t$ is

Accepted Answer

The electric field intensity ($E$) between the plates of a parallel plate capacitor is given by the formula:
$ \mathrm{E} = \frac{\mathrm{Q}}{\mathrm{A} \varepsilon} \quad 	ext{... (i)} $
Where:
$ Q $ is the charge on the capacitor.
$ A $ is the area of the plates.
$ \varepsilon $ is the permittivity of the dielectric material between the plates.
The capacitance ($C$) of a parallel plate capacitor is expressed as:
$ \mathrm{C} = \frac{\mathrm{A} \varepsilon}{\mathrm{t}} $
Solving for $ A \varepsilon $ gives:
$ A \varepsilon = \mathrm{Ct} \quad 	ext{... (ii)} $
Substituting equation (ii) into equation (i):
$ \mathrm{E} = \frac{\mathrm{Q}}{\mathrm{Ct}} $
Thus, the electric field intensity between the plates is $\frac{\mathrm{Q}}{\mathrm{Ct}}$.

Answer

$\frac{\mathrm{Qt}}{\mathrm{C}}$

Answer

$\frac{\mathrm{Q}}{\mathrm{Ct}}$

Answer

$\frac{\mathrm{C}}{\mathrm{Qt}}$

Answer

$\frac{\mathrm{Ct}}{\mathrm{Q}}$

Electrostatics & Capacitance

Charge on a parallel plate capacitor of capacity C is Q , the electric field intensity between its two plates separated by a distance of t is

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