Consider a parallel-plate capacitor made up of two conducting plates with dimens

Question

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 35 mm times 21 mm. If die separation between the plates is 1.1 mm. what is the capacitance, in pF, between them? Numeric: A numeric value is expected and not an expression. C = ______ If there is 044 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? Numeric: A numeric value is expected and not an expression. V = ______ What is the magnitude of die electric field, in newtons per coulomb, inside this capacitor? Numeric: A numeric value is expected and not an expression. E = ________ If die separation between the plates doubles, what will the electric field be if die charge is kept constant? Numeric: A numeric value is expected and not an expression. E' = ________

Explanation / Answer

Given that

area A=35*21=735*10^-6 m^2

distance d=1.1*10^-3 m

now we find the capacitance of capacitor

capacitance C=8.85*10^-12*735*10^-6/1.1*10^-3=5913.41*10^-15 F

now we find the voltage on capacitor

voltage V=0.44*10^-9/5913.41*10^-15=74.4 v

now we find the electric field

electric field E=74.4/1.1*10^-3=67.64*10^3 N/c

now we find the electric field

the electric field E=KQ/d^2

here we observed the electric field is directly proportional to charge and inversely propertional to distance

so if the charge is constant distance does not constant

so E1/E2=d2^2/d1^2

E1/E2=(2d)^2/(d)^2

E1/E2=4d^2/d^2

E2=e1/4=67.64/4*10^3=16.91*10^3 N/c

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