A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plate is 5 cm2, and the distance between the two plates is 1 mm. Assume the infinite plane approximation. 1) If we fix the charge on the capacitor to be Q = 6×10-11 C, what is the potential difference between the top and bottom plates? Note that in the region containing the dielectric medium, the electric field is E = E0 / κ, where E0 is the electric field in vacuum.

The value is

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is

The distance between the plates is

The charge on the capacitor is

Generally the electric field in a vacuum is mathematically represented as

Generally is the voltage of the capacitor which is mathematically represented as

Here is the capacitance of the capacitor in a vacuum which is mathematically represented as

Here is a constant with value

=>

=>

So

So

=>

The electric field when the dielectric slab is inserted is mathematically represented as

=>

=>

Generally the electric field between the plates is mathematically evaluated as

=>

=>

Generally the potential difference between the plates is

=>

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle