Question

In a parallel plate capacitor with air between the plates,each plate has an area 8×10^-3 m^2and distance between the plates is 2mm. a) calculated the capacitance of free space . b)if this capacitor is connected to a 50 V supply, what is the charge on each plate of the capacitor.

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Answer 1

Answer:

Q=1.771×10^(−9)

Explanation:

Given A=8×10−3m2

d=2×10−3m  

V=50V, e0=8.85×$10^{-2}$Fm−1 WKT electrical capacitance of an capacitor

C= ε0A / d

i.e, C= $\frac{(8.85.10^{-2}. 8.10^{-3}) }{ (2.10^{-3} )}$

i.e, C= 35.416×$10^{-2}$F (farad)

Also, WKT Q = Charge an the capacitor = CV

i.e, Q=35.416×$10^{-12}$×50

=1770.8×$10^{-12}$C

Q=1.771×$10^{-9}$(coulomb)

Answer 2

Answer:

A parallel plate capacitor with an area $8*10^{-3}m^{2}$  and separated by a distance of 2 mm the capacitance is  $352*10^{-13}F$, the charge stored when a $50v$ battery connected is $176*10^{-11}C$

Explanation:

• The parallel plate capacitor is two parallel plates that are connected across a battery when the plates are charged an electric field will induce between them.

• The CAPACITANCE is the ability to store electric charge within the capacitor

• For a parallel plate capacitor with an area $A$ separated by a distance $d$ in free space, the capacitance is

                                  $C=\frac{\varepsilon _0\ A}{d}$

• The amount of charge $Q$ that is stored in a capacitor of capacitance $C$ when a battery of voltage $V$  connected

                                  $Q=CV$                    

from the question, we have

area of square plate $A=8*10^{-3}m^{2}$

distance between two parallel plates $d=2mm=0.002m$

voltage  given $V=50v$

capacitance in free space $C=\frac{8.8*10^{-12} *8*10^{-3} }{0.002} =352*10^{-13} F$

the charge stored in a plate $Q=352*10^{-13}*50=176*10^{-11}C$