A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half

A parallel-plate capacitor is made from two plates 12.0 cm...

A parallel-plate capacitor is made from two plates 12.0 $\mathrm{cm}$ on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas\oplus of dielectric constant 3.40 . (See Figure $18.55 .$ An 18.0 V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel? (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas@, but change nothing else, how much energy will be stored in the capacitor?

Key Concepts

Effect of Removing a Dielectric in a Battery-Connected Circuit

When a dielectric is removed from a capacitor that remains connected to a constant voltage source, the capacitance changes while the voltage remains fixed. This change affects the stored energy, since the energy is directly proportional to the capacitance when the voltage is constant. The analysis of such changes is essential for understanding energy dynamics in practical capacitor applications.

Combination of Capacitors in Parallel

When multiple capacitors are connected in parallel, they all experience the same potential difference, and their individual capacitances add together to form the equivalent capacitance. This concept is used when a capacitor is constructed from different regions with varying materials (i.e., with and without a dielectric), where each region can be thought of as a separate capacitor connected in parallel.

Energy Stored in a Capacitor

The energy stored in a capacitor is given by the formula U = 1/2 C V^2, where C is the capacitance and V is the voltage applied across the capacitor. This formula illustrates how both the capacitor's ability to store charge and the applied voltage determine the total energy stored.

Parallel-Plate Capacitor

A parallel-plate capacitor is a fundamental electrical component that consists of two conductive plates separated by an insulating material. Its capacitance is determined by the area of the plates, the separation distance between them, and the permittivity of the medium between the plates. This basic model is widely used to understand more complex capacitor systems.

Dielectric Material

A dielectric material is an insulating substance placed between the plates of a capacitor. It increases the capacitor's ability to store charge by reducing the electric field for a given charge, effectively increasing the capacitance by a factor equal to the material's dielectric constant. Dielectrics are crucial for enhancing the performance of capacitors without changing their physical dimensions.

Transcript

00:01 In this question, we have a parallel plate capacitor that has square plates.

00:08 Then it has a separation of 4 .5mm.

00:13 What then happens is that we fill half of the space with a dialectic material with a constant of 3 .4.

00:21 And then we need to find out various properties of these capacitors.

00:26 So part a asks, what is the capacitance of this component? and the hint is to think about these capacitors as two capacitors in parallel.

00:38 Why? because we can have this configuration, but we can think of it as two capacitors, each with half the surface area of this one.

00:50 So this is 12 cm by 12 cm.

00:53 We will imagine this as 6th centimeter by 12 centimeters in the direction perpendicular to the paper.

01:00 And one of them having completely dialectic inner is completely filled with dialectic material.

01:09 And the other one is just air.

01:11 And then we can think of them as, we can think of them as, let me just remove this number.

01:22 Two capacitors that are in parallel.

01:30 Okay.

01:32 So, let's first thing, let's first try to find out their capacity.

01:37 Separately we have c equals k times epsilon times a over d so um let's call one of them c1 and call one of them c2 so let's say c1 is one with dialectic and c2 does not then we have c1 equals um c1 equals k abseal 0 a over d where k is 3 .4 and c2 is just epsilon not a over d because it has air and for air k equals 1 is a total capacitor c total is c1 plus c2 equals k minus 1 a k plus 1 times epsilon a over d so this is 3 .40 plus 1 times epsilon is 8 .854 times 10 to the negative 12 times the area is only one half the original one so it's 12 cm by 6 centimeter divided by 4 .50 times 10 to the negative 3 meter that's a separation this gives us 6 .25 times 10 to the negative 11 fahrenheit, or 62 .5 picofarad.

03:25 Picofarrant, sorry, not fahrenheit.

03:29 So the capacitors of this combination is 62 .5 picofarad.

03:35 And the next thing we need to find is we want to know the energy stored in this capacitor.

03:40 So for that, we just have u equals one half of cv square, because again, they share the same voltage.

03:48 So it's one half times 6 .25 times 10 to the negative 11 ferret times 18 .0 volt square...

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