Question

(a) Determine the capacitance of an isolated conducting sphere of radius \(a\) in free space (consider an outer conductor existing at \(r \rightarrow \infty\) ). \((b)\) The sphere is to be covered with a dielectric layer of thickness \(d\) and dielectric contant \(\epsilon_{r}\). If \(\epsilon_{r}=3\), find \(d\) in terms of \(a\) such that the capacitance is twice that of \(\operatorname{part}(a) .\)

Short answer

Answer: For the capacitance to be twice the original value, the thickness (d) of the dielectric layer must be equal to twice the radius of the sphere (a). Hence, the thickness of the dielectric layer should be d = 2a.

Step-by-step solution

Calculate the capacitance of an isolated conducting sphere

To find the capacitance of the sphere, use the formula for capacitance of a sphere: $$ C = 4 \pi \epsilon_0 a $$ where \(\epsilon_0\) is the permittivity of free space (approximately \(8.854 \times 10^{-12} \text{F/m}\)) and \(a\) is the radius of the sphere.

Calculate the capacitance of the sphere with the dielectric layer

Since the sphere is covered with a dielectric layer of thickness \(d\) and dielectric constant \(\epsilon_r\), the capacitance of the sphere with the dielectric layer can be found using the formula: $$ C' = 4 \pi \epsilon_0 a \epsilon_{r} $$

Set the capacitance with the dielectric layer to be twice the original capacitance

According to the exercise, the capacitance with the dielectric layer must be twice the original capacitance: $$ C' = 2C $$ Substituting the expressions for \(C\) and \(C'\) from steps 1 and 2, we get: $$ 4 \pi \epsilon_0 a \epsilon_{r} = 2(4 \pi \epsilon_0 a) $$

Solve for the thickness \(d\) in terms of \(a\)

Since the sphere with the dielectric layer can be treated as two concentric spheres with dielectric constant \(\epsilon_{r}\), the effective radius \(a + d\) of the whole sphere can be found as: $$ a + d = a \epsilon_{r} $$ Rearrange the equation to get: $$ d = a(\epsilon_{r} - 1) $$ Now substitute the given value of \(\epsilon_{r} = 3\): $$ d = a(3 - 1) $$ So, $$ d = 2a $$ Consequently, the thickness of the dielectric layer must be equal to twice the radius of the sphere for the capacitance to be twice the original value.

Key concepts

Electromagnetic Theory and Capacitance

Electromagnetic theory is a fundamental principle that describes how electric and magnetic fields interact with physical objects and vary in space and time. A key concept within this broad domain is the capacitance of conductive objects, which is essentially their ability to hold an electric charge.

When it comes to a conducting sphere, the capacitance is derived from the relationship between the electric field generated around the sphere when it is charged and the electric potential at its surface. The formula used in the exercise, \( C = 4 \pi \epsilon_0 a \), stems directly from this theory, where \( C \) is the capacitance, \( \epsilon_0 \) is the electric permittivity of free space, and \( a \) is the radius of the sphere. It's worth noting that this formula assumes there is another conductor at infinity, which essentially serves as a return path for electric field lines.

In electromagnetic theory, the electric permittivity of the medium surrounding the conductive object has a pivotal role. The higher the permittivity, the more the medium can polarize in response to an electric field, effectively allowing the conductor to store more charge. This directly influences the capacitance, as seen when we introduce dielectric materials into the equation.

Electric Permittivity in Electromagnetic Theory

Electric permittivity, often denoted by \( \epsilon \), is a measure of how an electric field affects, and is affected by, a dielectric medium. In the context of our isolated conducting sphere, \( \epsilon_0 \) represents the permittivity of free space and is a fundamental constant approximately equal to \(8.854 \times 10^{-12} \text{F/m}\). It sets the stage for understanding how electric fields interact with vacuum.

However, when materials are present, the concept of relative permittivity or dielectric constant \( \epsilon_r \) comes into play. It indicates how much a material can attenuate the electric field compared to the vacuum. For instance, if \( \epsilon_r \) is greater than 1, the field will get weaker. The capacitance of a conducting sphere increases linearly with \( \epsilon_r \), as shown in the step-by-step solution, where the presence of a dielectric material around the sphere directly multiplies the sphere's original capacitance by the dielectric constant, a concept that is an essential part of electromagnetic theory.

Dielectric Materials and Capacitance Enhancement

Dielectric materials are insulators that can be polarized by an electric field. They play a vital role in enhancing the capacitance of a conductor when used as a layer around it, as they effectively store electric charge. When a dielectric is introduced around our conducting sphere, it increases the capacitance by reducing the effective electric field and thus allowing more charges to accumulate on the sphere's surface for the same potential.

The revised formula for the capacitance of the sphere with a dielectric layer is \( C' = 4 \pi \epsilon_0 a \epsilon_{r} \), hinting at the direct proportionality between capacitance and the dielectric constant \( \epsilon_r \). As demonstrated in the solution process, to double the capacitance of the sphere, a dielectric layer with a relative permittivity of 3 (\( \epsilon_r = 3 \) in our case) leads to a layer thickness of twice the sphere's radius (\( d = 2a \) when \( \epsilon_r = 3 \) is substituted). This functional relationship is central to designing capacitors in practical applications, where dielectrics are chosen based on the desired electric properties and the need to control the device's capacitance.