# Parallel plate capacitor with dielectric. Find capacitance

• physics16102
In summary, a parallel plate capacitor with plates measuring 15cm x 15cm and separated by 6cm is charged with a potential difference of 18V. After a 15cm x 15cm x 3cm slab of dielectric material with a dielectric constant of 2.5 is inserted between the plates, the capacitance increases by a factor of 2.5. The final capacitance can be calculated using the formula C = εο*A /[(d-x) + x/k], where x is the thickness of the dielectric slab.
physics16102

## Homework Statement

The plates of a parallel plate capacitor measure 15cm x 15cm, and they are separated by 6cm. The capacitor is charged with a potential difference of 18V, then removed from the voltage source and kept isolated. A 15cm x 15cm x 3cm slab of dielectric material (k = 2.5) is inserted between the plates. What is the capacitance in µF after the dielectric is inserted?

Vi=Ei*d (1)
E=(ke)Q/r^2 (2)
V=(Vi)/k (3)
C=Q/V (4)

## The Attempt at a Solution

I used the formulas above in that order. I solved first for E(initial) using V(initial) and d=.06. I got E(i)=300
I then plugged E(i) it into eqn(2) to get Q. Q=1.2 x 10^-10
Next, I found V using eqn(3): 18V/2.5 to get V=7.2

Finally, I plugged Q and V into eqn 4 and got 1.7 x 10^-11

This doesn't match any of my multiple choice answers. Can someone help me figure out where I went wrong or give me a better plan of attack for this type of problem?

What is the capacitance in µF after the dielectric is inserted?

Capacitance does not depend on the charge on the plate or PD across the plates. When you introduce dielectric slab, the capacitance increases.

C = k*Co where C is the capacitance with dielectric, Co is the capacitance without dielectric and k is th dielectric constant.

Does the size of the dielectric affect the answer?
C for parallel plates= (eps)A/d
I used that to get C(i) and then multiply by the dielectric coefficient, but I am still incorrect

physics16102 said:
Does the size of the dielectric affect the answer?
C for parallel plates= (eps)A/d
I used that to get C(i) and then multiply by the dielectric coefficient, but I am still incorrect
If the dielectric is completely occupies the space between the plates, then the C = Co*k applies. If the a part of the space is filled, them

C = εο*A /[(d-x) + x/k], where x is the thickness of the dielectric slab.

Your approach seems correct, but the issue may lie in your calculations. Make sure you are using the correct units for each variable and double check your conversions. It may also be helpful to show your work so that any errors can be identified. Additionally, make sure you are using the correct formula for capacitance (C=Q/V) and not the one for voltage (V=Q/C). If you are still having trouble, try breaking down the problem into smaller steps and double check each calculation along the way.

## What is a parallel plate capacitor with dielectric?

A parallel plate capacitor with dielectric is a type of capacitor that consists of two parallel conductors separated by a non-conductive material called a dielectric. The dielectric material increases the capacitance of the capacitor by reducing the electric field between the two conductors.

## How does a parallel plate capacitor with dielectric work?

A parallel plate capacitor with dielectric works by storing electric charge on its plates. When a voltage is applied across the capacitor, positive and negative charges accumulate on the respective plates, creating an electric field between them. The dielectric material reduces this electric field, allowing for a larger charge to be stored on the plates.

## What is the capacitance of a parallel plate capacitor with dielectric?

The capacitance of a parallel plate capacitor with dielectric can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. The higher the permittivity of the dielectric, the larger the capacitance of the capacitor.

## How does the dielectric material affect the capacitance of a parallel plate capacitor?

The dielectric material affects the capacitance of a parallel plate capacitor by reducing the electric field between the plates. This allows for a larger charge to be stored on the plates, resulting in a higher capacitance. The permittivity of the dielectric also plays a role in determining the capacitance, as materials with higher permittivity can store more charge and therefore have a higher capacitance.

## What are some common materials used as dielectrics in parallel plate capacitors?

Some common materials used as dielectrics in parallel plate capacitors include air, plastic, ceramic, and paper. Each of these materials has different permittivity values, which affect the capacitance of the capacitor. The choice of dielectric material depends on the specific application and the desired capacitance value.

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