Question

A capacitor is constructed with two parallel metal plates each with an area of 0.55 m2 and separated byd = 0.80 cm. The two plates are connected to a 5.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.Find the electric field in the region between the two plates.Find the charge Q.Find the capacitance of the parallel plates.

Answer 1

To solve this problem it is necessary to apply the concepts related to energy, defined as the voltage by the distance of the seperation. Later we will find the load, as a function of voltage and capacitance. And finally we will find the capacitance between the two plates depending on the permittivity constant, the Area and the distance.

PART A) Electric field between two plates is

$E= \frac{V}{d}$

Here,

V = Voltage

d = Distance

Replacing,

$E = \frac{5.0V}{0.8*10^{-2}m}$

$E = 625V/m$

PART B) Charge on the capacitor plates is

$Q =CV$

Here,

C = Capacitance

V = Voltage

Replacing,

$Q = (5.75*10^{-10}F)(5.0V)$

$Q = 2.9*10^{-9}C$

PART C) Capacitance is

$C = \frac{\epsilon_0 A}{d}$

$\epsilon_0$= Permittivity of Free Space

A = Area

d = Distance

$C = \frac{(8.85*10^{-12}C^2/N\cdot m^2)(0.52m^2)}{0.8*10^{-2}m}$

$C = 0.00058*10^{-6}F$