Question

A parallel plate capacitor consists of square plates of edge length \(2.00 \mathrm{~cm}\) separated by a distance of \(1.00 \mathrm{~mm}\). The capacitor is charged with a \(15.0-\mathrm{V}\) battery, and the battery is then removed. A \(1.00-\mathrm{mm}\) -thick sheet of nylon (dielectric constant \(=3.0\) ) is slid between the plates. What is the average force (magnitude and direction) on the nylon sheet as it is inserted into the capacitor?

Short answer

The average force acting on the dielectric is 2.65 x 10^-6 N, and the direction of the force is towards the center of the capacitor (attractive).

Step-by-step solution

Capacitance without dielectric

Before the dielectric is inserted, we can find the capacitance of the parallel plate capacitor using the formula: \(C = \frac{\epsilon_0 A}{d}\), where \(C\) is the capacitance, \(\epsilon_0 = 8.85\times 10^{-12} F/m\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates. The area of a square plate of edge length \(2.00\,\text{cm}\) is \(A = (2.00 \times 10^{-2}\,\text{m})^2\). Given that the distance between plates is \(1.00\,\text{mm} = 1.00\times10^{-3}\,\text{m}\), we have: \(C = \frac{(8.85\times 10^{-12} F/m)((2.00 \times 10^{-2}\,\text{m})^2)}{(1.00\times10^{-3}\,\text{m})}=\boxed{3.54\times 10^{-11} \text{ F}}\).

Electric field and potential energy without dielectric

To find the electric field (\(E\)) without the dielectric, we can use the formula: \(E=\frac{V}{d}\), where \(V=15.0\,\text{V}\) is the voltage applied to the capacitor. So we have: \(E=\frac {15.0\,\text{V}}{1.00\times10^{-3}\,\text{m}}=\boxed{15.0\times10^3\,\text{V/m}}\). Next, we can find the potential energy (\(U\)) stored in the capacitor without the dielectric using: \(U = \frac{1}{2}CV^2\). By plugging in the capacitance and voltage, we get: \(U = \frac{1}{2}(3.54\times10^{-11}\,\text{F})(15.0\,\text{V})^2 = \boxed{3.98\times10^{-9}\,\text{J}}\).

Capacitance with dielectric

Now let's find the capacitance once the dielectric is inserted. We can use the formula \(C' = kC\), where \(C'\) is the capacitance with the dielectric and \(k = 3.0\) is the dielectric constant. Substituting our previous result for \(C\), we have: \(C' = (3.0)(3.54\times 10^{-11}\,\text{F}) = \boxed{10.6\times 10^{-11}\,\text{F}}\).

Electric field and potential energy with dielectric

The electric field with the dielectric is given by \(E' = \frac{E}{k}\). So, we can calculate \(E'\) as follows: \(E' = \frac{15.0\times10^3\,\text{V/m}}{3.0} = \boxed{5.00\times10^3\,\text{V/m}}\). Now we can find the potential energy with the dielectric as \(U' = \frac{1}{2}C'V'^2\). To find \(V'\), we can write \(Q = C'V'\), where \(Q\) is the charge stored on the capacitor. Since the battery has been removed, the charge remains constant. Thus we have \(Q = CV = C'V'\). Substituting our values and solving for \(V'\), we find \(V' = \frac{C}{C'}V = \frac{1}{3}V=\frac{1}{3}(15.0\,\text{V})=5.00\,\text{V}\). The potential energy with dielectric is then: \(U' = \frac{1}{2}(10.6\times10^{-11}\,\text{F})(5.00\,\text{V})^2 = \boxed{1.33\times10^{-9}\,\text{J}}\).

Calculate the work done on the dielectric and average force

The work done (\(W\)) on the dielectric during its insertion is equal to the change in potential energy: \(W = U - U'\). Substituting the values calculated above, we get: \(W = (3.98\times10^{-9}\text{J})-(1.33\times10^{-9}\,\text{J}) = 2.65\times 10^{-9}\,\text{J}\). Finally, we can calculate the average force (\(F\)) acting on the nylon sheet during its insertion using the formula: \(F = \frac{W}{d}\), where \(d = 1.00\,\text{mm}=1.00\times10^{-3}\,\text{m}\) is the distance of insertion. We find the force to be: \(F = \frac{2.65\times 10^{-9}\,\text{J}}{1.00\times10^{-3}\,\text{m}} = \boxed{2.65\times 10^{-6}\,\text{N}}\). The direction of the force is towards the center of the capacitor (attractive), since the system's potential energy decreases when the dielectric is inserted.

Key concepts

Capacitance With Dielectric

Capacitance measures a capacitor's ability to store electric charge. The presence of a dielectric material between the plates of a parallel plate capacitor affects its capacitance significantly. Describing this and enhancing the explanation of how a dielectric increases the capacitance is key.

A dielectric is an insulating material that, when placed between the plates of a capacitor, increases its capacitance by reducing the capacitor's electric field. This happens because the dielectric material becomes polarized in the presence of an electric field, aligning tiny electric dipoles within the material opposite to the field's direction. This internal field opposes the capacitor's original field, resulting in a net reduced electric field within the plates.

The increase in capacitance can be mathematically expressed as:
\(C' = kC\), where \(C'\) is the new capacitance with the dielectric, \(k\) is the dielectric constant (a measurement of the dielectric's effect on the electric field), and \(C\) is the original capacitance without the dielectric. A higher dielectric constant means a greater ability to store electrical energy for the same amount of electric field strength, explaining why the capacitance with the dielectric is multiplied by the dielectric constant.

Electric Field in Capacitors

The electric field within a parallel plate capacitor is a critical concept to understand when examining the behavior of capacitors. It is uniform between the plates and directly related to both the charge on the plates and the distance between them.

Without a dielectric, the electric field (\(E\)) is determined by the formula:\(E = \frac{V}{d}\), where \(V\) is the voltage across the plates and \(d\) is the separation between them. The effect of a dielectric can be understood as a reduction of this field. When a dielectric is introduced, the electric field (\(E'\)) with the dielectric becomes:\(E' = \frac{E}{k}\).

This decrease in the electric field reflects the physical concept that the dielectric opposes the field created by the separated charges on the capacitor plates. In simple terms, the dielectric 'pushes back' against the electric field, thereby allowing the capacitor to hold the same charge at a lower voltage or a higher charge at the same voltage.

Work Done on Dielectric

Understanding how work is done by the electric field when a dielectric is inserted into a capacitor adds a practical dimension to the theoretical concepts. To insert the dielectric, work must be done against the electric field within the capacitor. This work manifest as a decrease in the system's potential energy as the dielectric goes into the capacitor.

The work done (\(W\)) on the dielectric is given by the change in electric potential energy of the capacitor system:\(W = U - U'\), where \(U\) is the potential energy without the dielectric and \(U'\) is the potential energy with the dielectric present. This work is necessary to overcome the attraction between the dielectric and the electric field.

The insertion causes an attractive force towards the center of the plates. This is because inserting the dielectric reduces the system's potential energy—nature favors lower energy states, thus the electric field will act to pull the dielectric into the space where it can most effectively decrease the field. In quantitative terms, the average force (\(F\)) exerted on the dielectric can be calculated through the work done over the distance the dielectric is moved:\(F = \frac{W}{d}\). This would inform students about the relation between work, force, and energy in the context of electric fields and dielectric materials.