A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 c

step 1 In this question, we have a parallel plate capacitor in air. Okay, so it has a plate separation

A parallel-plate capacitor in air has a plate separation of...

A parallel-plate capacitor in air has a plate separation of $1.50 \mathrm{~cm}$ and a plate area of $25.0 \mathrm{~cm}^{2}$. The plates are charged to a potential difference of $2.50 \times 10^{2} \mathrm{~V}$ and disconnected from the source. The capacitor is then immersed in distilled water. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy stored in the capacitor due to immersion. Assume the distilled water is an insulator.

A parallel-plate capacitor in air has a plate separation of $1.50 \mathrm{~cm}$ and a plate area of $25.0 \mathrm{~cm}^{2}$. The plates are charged to a potential difference of $2.50 \times 10^{2} \mathrm{~V}$ and disconnected from the source. The capacitor is then immersed in distilled water. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and
(c) the change in energy stored in the capacitor due to immersion. Assume the distilled water is an insulator.

Key Concepts

Energy Stored in a Capacitor

The energy stored in a capacitor is given by U = ½CV² or alternatively U = Q²/(2C). When a dielectric is introduced, the shift in capacitance and potential difference directly influences the stored energy. For a capacitor disconnected from a power source, the conservation of charge and the resulting change in voltage lead to a reduction in stored energy, which is an important factor in analyzing the overall energy profile of the system.

Capacitance with Dielectric Versus Air

The introduction of a dielectric material such as distilled water increases the capacitance by a factor equal to the dielectric constant. The new capacitance is given by C' = k??A/d, where k is the dielectric constant. This increased capacitance, under constant charge conditions in an isolated capacitor, leads to a decrease in the potential difference across the plates.

Dielectric Materials

Dielectrics are insulating materials that, when placed between the plates of a capacitor, increase its capacitance by reducing the effective electric field. This is quantified by the dielectric constant (relative permittivity) of the material, which multiplies the capacitance when compared to a vacuum or air. The insertion of a dielectric changes the voltage across the capacitor, even though the charge may remain constant if the capacitor is isolated.

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by a distance, where the capacitance is determined by the area of the plates, the separation between them, and the permittivity of the medium between the plates. The basic formula without a dielectric is C = ??A/d, which describes how the physical dimensions and material properties affect the ability to store electric charge.

Charge Conservation in an Isolated Capacitor

When a capacitor is disconnected from the charging source, it becomes isolated, and the charge stored on its plates remains constant regardless of changes to other system parameters. This concept is crucial when analyzing changes in potential difference and energy storage upon the introduction of a dielectric, as any alteration in capacitance must occur at the expense of the potential difference under constant charge.

Transcript

00:01 In this question, we have a parallel plate capacitor in air.

00:05 Okay, so it has a plate separation of 1 .5 cm.

00:14 The area is 25 cm square.

00:20 And then it is charged to 250 volts.

00:27 After that, it's disconnected, then it's immersed in distilled water.

00:31 So we want to determine the charge on the plates before and after immersion, capacitance and potential difference after emergent, and the change in energy stored in the ground.

00:40 Capacitor due to immersion.

00:42 So in part a, to find a charge on the capacitor before and after the immersion.

00:52 So this is how we do it.

00:57 Okay, since the capacitor is disconnected after before immersion, so the charge, since the capacitor is disconnected from the source before immersion, so the charge before and after are the same is equal to cv.

01:42 So c0b and c0 is epsilon a over d times b.

01:52 So we just put in the numbers, 8 .85 times 10 ,000 part negative 12 times 25 times 10 to pound negative 4, divide by 1 .5 times 10 to power negative 2, multiplied by 2050.

02:06 So the charge stored on the plate is 3 .69 times 10 to the power negative 10 columns.

02:16 Okay, so this is the answer for part a.

02:23 Then for part b to find a potential, the capacitance and potential difference after immersion.

02:30 Okay, so before immersion.

02:41 Okay, the capacitance, to find the capacitance after immersion, so c is equal to kappa c0, which is kappa epsilon not a over d.

03:01 So kappa is ad for distilled water, and then 8 .85 times 10 to pound negative 12, times 25 times 10 to pound negative 4, divide by 1 .5 times 10 to the pound negative 2, and the answer is 1 .18 times 10 to the power negative 10 ferret...

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