Answer
$E = 1.75 \times 10^{5} V/m $
Work Step by Step
Given:
Consider a parallel-plate capacitor having an area of 3225 $mm^2$ (5 $in.^2$), a plate separation of 1 mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates.
Required:
electric field that must be applied for $2 \times 10^{-8}$ C to be stored on each plate
Solution:
Using a combination of Equation18.24 and the value of C from Item 18.51a, we solve for V:
$V = \frac{Q}{C} = \frac{2 \times 10^{-8}}{1.14 \times 10^{-10} F} = 175.44 V $
Using Equation 18.6:
$E = \frac{V}{l} = \frac{175.44 V}{(1.0 \times 10^{-3} m)} = 1.75 \times 10^{5} V/m $
