An electric field of 7.00×10^5 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?

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Answer 1 · 2018-02-27T16:25:26

#sf(2.79xx10^(-8)color(white)(x)C)#

The capacitance is given by:

#sf(C=(epsilonA)/d)#

For air the relative permittivity =1.0006

#:.##sf(C=(8.85xx10^(-12)xx1.0006xx4.5xx10^(-3))/(2.45xx10^(-3))color(white)(x)F)#

#sf(C=16.26color(white)(x)pF)#

#sf(V/d=7.00xx10^(5)color(white)(x)"V/m")#

#:.##sf(V=7.00xx10^(5)xx2.45xx10^(-3)=1715color(white)(x)V)#

#sf(C=Q/V)#

#:.##sf(Q=CV=16.26xx10^(-12)xx1715=2.79xx10^(-8)color(white)(x)C)#