PART A & B MUST BE ANSWERED TO RECEIVE FULL CREDIT (AND CORRECT): Take your time

Question

PART A & B MUST BE ANSWERED TO RECEIVE FULL CREDIT (AND CORRECT): Take your time.

PART A:

In our lab we used an air filled capacitor. However, in most applications, capacitors are filled with a dielectric, non-conducting materials, that increase the capacitance. If the dielectric constant of air is 1, then in order to increase the capacitance, then the dielectric constant of a dielectric filled capacitor must have a value that is:

equal to 1

greater than 1

less than 1

Not enough information

PART B:

In this experiment, the capacitance of the plates are defined by C equal fraction numerator italic epsilon subscript o A over denominator d end fraction , where A is the ares of the plates, d is the separation of the plates, and italic epsilon subscript o is the permittivity of free space by an electric field. What happens to the capacitance when the side of the square plate is decreased by half and the distance is increased by 3?

The capacitance is increased by 12

The capacitance is increased by 6

The capacitance is decreased by 6

The capacitance is decreased by 12

Explanation / Answer

1. greater than 1  

C = KeoA/d where K is the value of di elecrric which enhances the the net cacpaciatnce

2. C= keoA/d

if side in decreased by two, area = a^2/4

and d'=3d

so Cnew = keoA/12d

Cnew = The capacitance is decreased by 12

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