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A metal sphere with radius \(r_{a}\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(\eta_{b}\). There is charge \(+q\) on the inner sphere and charge \(-q\) on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(rn\). (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$ V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{n_{b}}\right) $$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude \(E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}\) (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r>r_{b} .\) (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q .\) Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different

Step by step solution

01

Calculating the Potential Inside the sphere (\(r

Inside the charged sphere, the electric field is zero. Since the electric field is the negative gradient of potential, when E is zero, V is constant. So, for \(r

02

Calculating the Potential Outside the sphere (\(r_{a}

Between the two spheres, the only charge within the sphere of radius \(r\) is \(+q\). So, \(V(r)-V_{a}=-\int_{{r_{a}}}^{r}E.dr\), where \(E= \frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\) from Gauss's Law. Integrating gives \(V_{a}-V(r)=\frac{q}{4\pi\epsilon_{0}}(\frac{1}{r}-\frac{1}{r_{a}})\). Therefore, \(V(r)= V_{a}+\frac{q}{4\pi\epsilon_{0}}(\frac{1}{r_{a}}-\frac{1}{r})\)

03

Calculating the Potential Beyond the sphere (\(r>n_{b}\))

For \(r>n_{b}\), the potential is given by \(V(r)=V_{b}+\frac{q}{4\pi\epsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r})\), Since at \(r\rightarrow \infty\), \(V=0\), \(V_{b}\) is equal to \(-\frac{q}{4\pi\epsilon_{0}}\frac{1}{r_{b}}\). Thus, \(V(r)=\frac{q}{4\pi\epsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r}) - \frac{q}{4\pi\epsilon_{0}}\frac{1}{r_{b}} = -\frac{q}{4\pi\epsilon_{0}}\frac{1}{r}\)

04

Calculating the Potential Difference \(V_{a b}\)

The potential difference between the two spheres is \(V_{a b}=V_A - V_B = \frac{q}{4\pi\epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{r_{b}}\right)\)

05

Calculating the Electric Field Between the Spheres

Using the relation \(E=-dV/dr\), The electric field at any point between the spheres is \(E(r)=\frac{V_{ab}}{(1/r_{a}-1/r_{b})}\frac{1}{r^{2}}\)

06

Calculating the effect of charge -Q on the outer sphere

If the outer sphere has a charge of \(-Q\), the net charge enclosed between the two spheres still remains \(+q\). Therefore, the answers for \(V_{ab}\) and \(E(r)\) do not change. But, for \(r>n_{b}\), \(V(r)=-\frac{Q}{4\pi\epsilon_{0}}\frac{1}{r}\) and using \(E=-dV/dr\),so\(E(r)=\frac{Q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential

Electric potential, often denoted as \( V \), is a measure of the potential energy per unit charge at a point in a field. It's essential to understand that the potential is related to the electric field by the equation \( E = -abla V \), where \( E \) is the electric field. This means the electric field is the negative gradient of the electric potential.
In the context of a spherical conductor, we must consider how the electric potential changes in different regions:

• Within the inner sphere: Here, the electric field is zero because the potential is constant. Hence, \( V(r) = V_a \), a constant.
• Between the spheres: The potential here is affected by charge \(+q\) from the inner sphere, leading to \( V(r) = V_a + \frac{q}{4\pi\epsilon_0}(\frac{1}{r_a} - \frac{1}{r}) \).
• Outside the outer sphere: As we move far away, the potential approaches zero since the effects of the charge diminish with distance.

Ultimately, understanding electric potential requires grasping how potential changes in space relative to static charges.

Spherical Conductors

Spherical conductors have a unique property where the charge resides only on their surfaces. This results because charges are mobile and will arrange themselves to maintain equilibrium. In such setups,
the electric field inside the conductor is zero. This absence of an internal field means that any excess charge distributes itself only on the surface, creating a shield effect. This phenomenon is why the potential remains constant inside the conductor.
For a system of spherical conductors, the charges on each adopt a configuration to minimize energy, leading to interesting behaviors in potentials and fields. When considering multiple conductive spheres, such as an inner charged sphere inside a larger one, the understanding of their interactive potentials becomes critical. This further explains why potential differences and field distributions behave as calculated in the exercise.

Electric Field

The electric field \( E \) represents the force a charge would experience per unit charge at any point in space. It's direction and magnitude are derived from the distribution of surrounding charges. Mathematically, it's denoted as \( E = -abla V \), indicating it flows in the direction where the potential decreases most rapidly.
For our spherical conductor setup:

• Between the two spheres: The electric field is given by \( E(r) = \frac{V_{ab}}{(1/r_a - 1/r_b)} \frac{1}{r^2} \). This expression reflects the potential difference \( V_{ab} \) between the spheres.
• Outside both spheres: The field can be modeled like that due to point charges at the center, diminishing in strength with the square of the distance \( r \), \( E(r) = \frac{Q}{4\pi\epsilon_0} \frac{1}{r^2} \).

This illustrates how electric field lines diffuse outward from charged surfaces and diminish over distance. Recognizing these patterns is crucial for predicting force interactions in physics.

Gauss's Law

Gauss's law is a pivotal principle in electrostatics. It states that the net flux of an electric field through a closed surface is directly proportional to the enclosed charge. Mathematically, it's represented as \( \Phi = \oint E \cdot dA = \frac{Q_{ ext{enc}}}{\epsilon_0} \).
In terms of spherical conductors, Gauss's law provides practical results:

• Inside a conductor, the net enclosed charge is zero, hence the electric field is zero.
• Between concentric conductors, only charges within the inner boundaries contribute to the field.
• Outside of a charged conductor, the field behaves as if it were emanating from a point charge located at the center of the sphere.

Gauss's law simplifies complex calculations, allowing us to deduce fields and potentials intuitively by focusing primarily on charge distribution and geometry.