Question

A parallel-plate capacitor has plates located at \(z=0\) and \(z=d\). The region between plates is filled with a material that contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\) and has permittivity \(\epsilon\). Both plates are held at ground potential. ( \(a\) ) Determine the potential field between plates. (b) Determine the electric field intensity \(\mathbf{E}\) between plates. \((c)\) Repeat parts \((a)\) and \((b)\) for the case of the plate at \(z=d\) raised to potential \(V_{0}\), with the \(z=0\) plate grounded.

Short answer

Question: Determine the following for the given parallel-plate capacitor: a) The potential field between the plates b) The electric field intensity between the plates c) The potential field and electric field intensity when the plate at \(z=d\) is raised to potential \(V_0\) Answer: a) The potential field between the plates is given by: $$ V(z) = \frac{\rho_0}{4\epsilon}z^2 $$ b) The electric field intensity between the plates is given by: $$ \mathbf{E} = -\frac{\rho_0}{2\epsilon}z\hat{z} $$ c) When the plate at \(z=d\) is raised to potential \(V_0\), the potential field and electric field intensity are given by: $$ V(z) = \frac{\rho_0}{4\epsilon}z^2 - \frac{V_0}{d}z $$ $$ \mathbf{E} = \left(\frac{\rho_0}{2\epsilon}z - \frac{V_0}{d}\right)\hat{z} $$

Step-by-step solution

Part (a): Determine the potential field between plates

Use Gauss's law to find the electric field between the plates: $$ \nabla\cdot\mathbf{D} = \rho_0 $$ where \(\mathbf{D} = \epsilon\mathbf{E}\) is the electric displacement. Since the permittivity of the material is \(\epsilon\), it follows that $$ \nabla\cdot\mathbf{E} = \frac{\rho_0}{\epsilon} $$ Integrating, we get $$ \mathbf{E} = -\frac{\rho_0}{2\epsilon}z\hat{z} $$ Now use the relation between the electric field and potential: $$ \mathbf{E} = -\nabla V $$ We now integrate the electric field to find the potential: $$ V(z) = -\int_0^z \mathbf{E}\cdot d\mathbf{l} = \frac{\rho_0}{4\epsilon}z^2 + C $$ To find the constant \(C\), use the boundary condition that both plates are at ground potential (\(V(0) = V(d) = 0\)): $$ C = 0 \Rightarrow V(z) = \frac{\rho_0}{4\epsilon}z^2 $$ So the potential field between the plates is $$ V(z) = \frac{\rho_0}{4\epsilon}z^2 $$

Part (b): Determine the electric field intensity between plates

From the previous step, we know the potential field between the plates, \(V(z)\). We can now find the electric field intensity \(\mathbf{E}\) by taking the gradient of the potential field: $$ \mathbf{E} = -\nabla V(z) = -\frac{\rho_0}{2\epsilon}z\hat{z} $$

Part (c): Repeat parts (a) and (b) for the case of the plate at \(z=d\) raised to potential \(V_{0}\)

When the plate at \(z=d\) is raised to potential \(V_0\), the potential field can be treated as a superposition of the previous case (with both plates grounded) and the new case (with the plate at \(z=d\) at potential \(V_0\) and the plate at \(z=0\) grounded). Let \(V_1(z)\) be the potential field for the grounded case and \(V_2(z)\) be the potential field for the new case. Then, the total potential field is given by $$ V(z) = V_1(z) + V_2(z) $$ The potential field for the grounded case, \(V_1(z)\), is given by $$ V_1(z) = \frac{\rho_0}{4\epsilon}z^2 $$ To find the potential field \(V_2(z)\) for the new case, note that it satisfies the Laplace equation \(\nabla^2 V_2(z) = 0\) since no new charges are added. Hence, we can write \(V_2(z)\) as $$ V_2(z) = Az + B $$ Using the boundary conditions \(V_2(0) = 0\) and \(V_2(d) = -V_0\), we get $$ B = 0, \quad A = -\frac{V_0}{d} $$ Thus, $$ V_2(z) = -\frac{V_0}{d}z $$ Adding the two potentials, we obtain $$ V(z) = \frac{\rho_0}{4\epsilon}z^2 - \frac{V_0}{d}z $$ Now, we can find the electric field intensity by taking the gradient of the total potential field $$ \mathbf{E} = -\nabla V(z) = \left(\frac{\rho_0}{2\epsilon}z - \frac{V_0}{d}\right)\hat{z} $$

Key concepts

Electric Field Intensity

The electric field intensity is a vector quantity that represents the force experienced per unit charge at a point in space. In the context of a parallel-plate capacitor filled with a dielectric material of permittivity \(\epsilon\), the electric field intensity is uniform between the plates and is directly proportional to the volume charge density \(\rho_{0}\).

According to the given step-by-step solution, the electric field intensity \(\mathbf{E}\) between the grounded plates can be derived from \(abla\cdot\mathbf{E} = \frac{\rho_0}{\epsilon}\). The electric field intensity is then integrated and simplified to \(\mathbf{E} = -\frac{\rho_0}{2\epsilon}z\hat{z}\).

Understanding Electric Field Intensity in Capacitors

To help with comprehension, picture the space between the capacitor's plates as being filled with field lines. These lines are evenly spaced and parallel to each other, indicating a uniform electric field. The vector nature of \(\mathbf{E}\) reveals the direction in which a positive charge would be pushed, which is away from the positively charged plate and toward the negatively charged plate.

The strength of the field \(\mathbf{E}\) is determined by the volume charge density; a higher \(\rho_{0}\) will result in a stronger electric field. This relationship is crucial for students to understand, as it governs how the electric field behaves within different materials and configurations.

Electric Potential Field

The electric potential field, commonly referred to as electric potential, is a scalar quantity that represents the work done to move a unit charge from a reference point (usually infinity) to a specific point in an electric field without acceleration. For the parallel-plate capacitor, the potential field is a function of the position between the plates and is derived by integrating the electric field intensity.

The potential field between the plates of a capacitor that contains a charge distribution is expressed by the formula \(V(z) = \frac{\rho_0}{4\epsilon}z^2\), where \(z\) is the distance from one of the plates. As shown in the solution, the boundary condition, that both plates are at ground potential, requires the constant of integration to be zero.

Visualizing Electric Potential in a Capacitor

The potential field, unlike the electric field intensity, does not have direction. You can visualize it as a landscape of hills and valleys where a positive charge would naturally 'roll down' from a region of high potential to low potential, similar to a ball rolling downhill due to gravity. It's important for students to understand that the potential difference between the plates is what drives the current when the circuit is complete, and the shape of the potential field inside the capacitor can affect its energy storage properties.

Gauss's Law

Gauss's Law is a fundamental equation of electromagnetism encapsulated in Maxwell's equations. It relates the electric field in a region of space to the charge present in that region. The law states that the net electric flux through any closed surface is equal to the charge enclosed divided by the permittivity of free space.

In the step-by-step solution provided, Gauss's Law is rendered as \(abla\cdot\mathbf{D} = \rho_0\), where \(\mathbf{D}\) is the electric displacement field related to the electric field \(\mathbf{E}\) and the material's permittivity \(\epsilon\). This is utilized to calculate the electric field and, subsequently, the potential field in a parallel-plate capacitor.

Applying Gauss's Law to a Capacitor

An intuitive way to understand Gauss's Law is to think of it as an accounting method for electric field lines. A closed surface around a charge will 'catch' a number of field lines proportional to the amount of charge. In our capacitor scenario, the law assists in determining the field created by the uniform charge density. By knowing the electric field, one can then find the influence on a test charge placed within the electric field. This understanding helps students visualize how charges within a configuration like a capacitor can influence the surrounding electric field.