# How Does Inserting a Dielectric Affect Capacitor Voltage?

• rcrx
In summary, when an empty capacitor is connected to a 12.0-V battery and then disconnected and a slab of dielectric material (K = 2.8) is inserted between the plates, the potential difference across the plates will change. The charge on the capacitor remains the same, but the voltage will drop. To calculate the change in potential difference, you can use the formula q = CV and the fact that the charge remains the same. The new capacitance (C2) can be calculated using the formula C2 = KC1, where C1 is the initial capacitance. Finally, the new potential difference can be found by using the formula V = q/C2.
rcrx

## Homework Statement

An empty capacitor is connected to a 12.0-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (K = 2.8) is inserted between the plates. Find the amount by which the potential difference across the plates changes.

C2 = KC1
q = CV

## The Attempt at a Solution

I know that the charge will stay the same and voltage will drop.

I have no idea how to set this up mathematically. This is where I have hit a road block.

You are not give C1, so you must think of it as a "known" that may appear in the answer. From that and one of your formulas, you can calculate the charge.

Great observation to see that the charge remains the same with the dialectric is inserted!

Use the formula with the V in it to find the potential on the cap with dielectric.

Thanks for the reply. But how do I set it up? I am so confused. I know it is simple, but some problems I just can't wrap my head around mathematically.

Any and all help is very much appreciated!

Well, to calculate the charge, I just meant to use your formula q = CV.
To avoid confusion, I would replace C with C1. Put in your potential and you've got an expression for q with no unknowns (except C1, which we are treating as a known for now).

The first formula gives the new C2.
Use the 2nd formula again to find the new V.

## What is capacitance?

Capacitance is the ability of a conductor to store electric charge. It is measured in Farads (F) and is dependent on the size and shape of the conductor, as well as the distance between the conductors.

## How is capacitance calculated?

The capacitance of a conductor is calculated by dividing the amount of charge stored on the conductor by the potential difference between the conductors. It can also be calculated by multiplying the permittivity of the material between the conductors by the area and dividing by the distance between the conductors.

## What is a dielectric material?

A dielectric material is an insulating material that is placed between the conductors of a capacitor. It is used to increase the capacitance of the capacitor by reducing the electric field between the conductors. Common dielectric materials include air, paper, plastic, and ceramic.

## How does a dielectric affect capacitance?

A dielectric material increases the capacitance of a capacitor by reducing the electric field between the conductors. This is because the dielectric material has a higher permittivity than air, which allows for more charge to be stored on the conductors.

## What is the role of capacitance in electronic circuits?

Capacitance plays a crucial role in electronic circuits as it allows for the storage of electric charge, which can be used for various purposes such as filtering, energy storage, and timing. It is also used in combination with other components to create filters, oscillators, and other circuit elements.

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