Question

You need to design a capacitor capable of storing 3.0 10-7 c of charge. at your disposal, you have a 100 v power supply and two metal plates, each of area 0.490 m2 each. what is the limit of the separation of the plates? mm

Answer 1

The capacitor formula is given by  
Capacitance = Q/V  
(3 * 10-7)/100 = 3*10-9  
the relation between capacitance , Area and distance between the plates is given by 
 C = (EoA)/d --> (EoA)/C = d  
(8.85 * 10-12) * 0.490 /3*10-9  
 d=.1.44mm

Answer 2

The separation between the plates of the capacitor is $1.445 \tiems 10^{-3} \;\rm m$.

What is the capacitor?

The capacitor is a device that is used to store charge. The capacity of a capacitor to store charge is called capacitance.

Given that the charge Q is 3.0 X 10^-7 C and the voltage V is 100 V. The cross-sectional area A of each plate of the capacitor is 0.490 m2.

The capacitance C is calculated as given below.

$C = \dfrac {Q}{V}$

$C = \dfrac { 3.0 \times 10^{-7}}{100}$

$C = 3.0 \times 10^{-9}$

The capacitance of the capacitor is given below.

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

Where d is the distance between the plates of the capacitor and $\epsilon _0$ is the permittivity of free space which is 8.85 x 10^-12.

$3.0 \times 10^{-9} = \dfrac {8.85\times 10^{-12} \times 0.490}{d}$

$d = \dfrac {8.85\times 10^{-12} \times 0.490}{3.0\times 10^{-9}}$

$d = 1.445 \times 10^{-3} \;\rm m$

Hence we can conclude that the separation between the plates of the capacitor is $1.445 \tiems 10^{-3} \;\rm m$.

To know more about the capacitor, follow the link given below.

https://brainly.com/question/14048432.