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If a capacitor has opposite 4.9 μc charges on the plates, and an electric field of 1.6 kv/mm is desired between the plates, what must each plate's area be?

Respuesta :

tatamaraweeks74
Q = 4.6E-6 C 
E = 2.6 kV/mm = 2600 kV/m = 2600000 V/m 
ε = 8.85E-12 F/m 
E = Q/Aε 
A = Q/Eε = 4.6E-6/(2600000*8.85E-12) = 0.2 m^2
okpalawalter8

We have that The area of the Capacitor

[tex]a=0.3*10^3m^2[/tex]

From the question we are told that

a capacitor has opposite 4.9 μc charges on the plates

an electric field of 1.6 kv/mm is desired between the plates

Generally the equation for the Area  is mathematically given as

[tex]a=\frac{Cd}{e_0}\\\\a=\frac{4.9*10^-{6}}{1.6*10^3*8.85*10^{-12}}[/tex]

[tex]a=0.3*10^3m^2[/tex]

Therefore

The area of the plate is

[tex]a=0.3*10^3m^2[/tex]

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