Answer :
Based on the problem, the distance between two parallel conducting plate is x=10cm=0.10mx=10cm=0.10m
The electric field
a)
The potential difference is V= 593VV= 593V
The distance between is d=7.05cm= 0.705md= 7.05cm= 0.705m
Recall the expression for the electric field.
E= VdE= Vd
Substitute the given values in above equation.
E= 5930.705= 841.13V/ mE =5930.705 = 841.13V/Vmm
Thus, the electric field strength is 841.13V/m.841.13V/Vmm.
b)
Recall the term for the voltage between the plate.
ΔV= E× x ΔV= E×x
Substitute the given values in above equation.
ΔV=841.13V/m×0.10m=84.11VΔV=841.13V/Vmm×0.10m=84.11V
Thus, the voltage within the plate is 84.11V.
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