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Chapter 19: Problem 24

Point charges \(q\) and \(-q a / b\) (with \(a

Short Answer

Expert verified
Net potential on the sphere is zero by similar triangles' symmetry. The force on the charge is \( \frac{q^2 a}{4\pi \epsilon_0 b(b-a)} \) in both earthed and insulated cases.

Step by step solution

01

- Define the Problem and Setup

We need to show that the net potential due to two charges, one at point P (distance b from origin O) and another at point Q (distance \(a^2/b\) from O), is zero on a sphere of radius a centered at O. We also need to find the force on a charge q placed at a distance b from this sphere.
02

- Understand the Placement of Charges

Charge \(q\) is placed at point P, at a distance b from the origin O. Another charge \(- \frac{q a}{b}\) is placed at point Q, at a distance \(\frac{a^2}{b}\) from the origin O.
03

- Define the Positions using Coordinates

Let the position vectors be \( \vec{r}_P = \vec{b} \) for point P and \( \vec{r}_Q = \frac{a^2}{b} \widehat{PQ} \) for point Q. The point on the surface of the sphere S has the position vector \( \vec{r}_S = a \widehat{SO} \).
04

- Apply the Similar Triangles

According to the problem, consider similar triangles \( QOS \) and \( SOP \). This implies the ratios of corresponding sides are equal: $$ \frac{OQ}{OS} = \frac{OS}{OP}, $$ or $$ \frac{\frac{a^2}{b}}{a} = \frac{a}{b}. $$
05

- Calculate the Potential from Each Charge

The potential due to a point charge is given by: $$ V = \frac{k q}{r}, $$ where k is Coulomb's constant and r is the distance to the point where potential is calculated. For point S on the sphere: $$ V_{P}(S) = \frac{k q}{|\vec{r}_S - \vec{b}|} $$ and $$ V_{Q}(S) = \frac{k \left( -\frac{q a}{b} \right)}{|\vec{r}_S - \frac{a^2}{b} \widehat{PQ}|}. $$
06

- Let us Write the Net Potential at Point S

The total potential at any point S on the sphere would be: $$ V_{net}(S) = V_{P}(S) + V_{Q}(S). $$ Given the similar triangles, \(|\vec{r}_S - \vec{b}| = b - a\) and \(|\vec{r}_S - \frac{a^2}{b} \widehat{PQ}| = a - \frac{a^2}{b} = \frac{a(a-b)}{b},\) which on symmetrical analysis confirms the net potential zero.
07

- Use the Uniqueness Theorem to Determine Force

According to the uniqueness theorem, if the potential distribution is uniquely zero on the spherical surface, then the field satisfying boundary conditions is unique. Calculate the image charge for finding the force.
08

- Consider Earth Scenario

For an earthed sphere, the potential is zero, and we calculate the attractive force as if there were an image charge inside: $$ F = \frac{q^2 a}{4\pi \epsilon_0 b(b-a)}. $$
09

- Consider Insulated Sphere

If the sphere is uncharged and insulated, the force calculation is the same as in earthed due to no other charge being present: $$ F = \frac{q^2 a}{4\pi \epsilon_0 b(b-a)}. $$

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Key Concepts

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

Coulomb's Law
Coulomb's Law forms the foundation of electrostatics. This law quantifies the force between two point charges. It states that the force between them is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where \(F\) is the magnitude of the force, \(q_1\) and \(q_2\) are the charges, \(r\) is the separation distance, and \(k_e\) is Coulomb's constant \(8.988 \, \times 10^9 \, \text{Nm}^2 \text{C}^{-2}\). Understanding this law helps us calculate electrical forces in various configurations, such as the case of two charges positioned as described in the exercise.
Potential Theory
Potential theory deals with the electric potential in an electrostatic field. The electric potential \(V\) at any point is the work done per unit charge to bring a small positive test charge from infinity to that point: \[ V = k_e \frac{q}{r} \] where \(q\) is the point charge and \(r\) is the distance from the charge to the point of interest. In the given exercise, we compute the potential at a point \(S\) on the sphere due to two different charges. By superimposing both potentials, we can determine if the net potential at that point is zero. This principle guides us in solving complex potential problems by breaking them down into simpler parts.
Image Charge Method
The image charge method is a useful tool in electrostatics to simplify problems involving conductors. This method replaces the actual problem with a more manageable one by introducing imaginary charges (image charges) that replicate the boundary conditions on the conductor’s surface. For the exercise, imagine an image charge inside the sphere that will ensure the boundary condition (zero potential on the sphere) is satisfied. For a charge \(q\) outside a conductor, the image charge concept lets us compute the forces as if the actual charges interact with these imaginary ones.
Uniqueness Theorem
The Uniqueness Theorem in electrostatics states that the potential in a given region is uniquely determined if the potential is specified on the boundary of the region. This theorem guarantees that the potential solution found is the only solution, making it a powerful tool in electrostatic problem-solving. In the exercise, after establishing that the net potential on the sphere is zero, we use the Uniqueness Theorem to confirm that the calculated potential distribution is unique. This leads to a unique calculation of the electrical forces involved, ensuring our solution's completeness and correctness.
Electrical Forces
Electrical forces arise between charged particles due to Coulomb's law. These forces can be attractive or repulsive depending on the signs of the interacting charges. In this exercise, we need to determine the force on a charge \(q\) placed outside a spherical conductor. We apply the result from potential theory and image charge method to get the force expression: \[ F = \frac{q^2 a}{4\pi \epsilon_0 b(b-a)} \] This formula is derived based on whether the sphere is earthed or insulated and confirms that the attractive force depends on the charge and the geometry of the system.

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Most popular questions from this chapter

The wave equation describing the transverse vibrations of a stretched membrane under tension \(T\) and having a uniform surface density \(\rho\) is $$ T\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right)=\rho \frac{\partial^{2} u}{\partial t^{2}} $$ Find a separable solution appropriate to a membrane stretched on a frame of length \(a\) and width \(b\), showing that the natural angular frequencies of such a membrane are $$ \omega^{2}=\frac{\pi^{2} T}{\rho}\left(\frac{n^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}\right) $$ where \(n\) and \(m\) are any positive integers.

The motion of a very viscous fluid in the two-dimensional (wedge) region \(-\alpha<\) \(\phi<\alpha\) can be described in ( \(\rho, \phi\) ) coordinates by the (biharmonic) equation $$ \nabla^{2} \nabla^{2} \psi \equiv \nabla^{4} \psi=0 $$ together with the boundary conditions \(\partial \psi / \partial \phi=0\) at \(\phi=\pm \alpha\), which represents the fact that there is no radial fluid velocity close to either of the bounding walls because of the viscosity, and \(\partial \psi / \partial \rho=\pm \rho\) at \(\phi=\pm \alpha\), which imposes the condition that azimuthal flow increases linearly with \(r\) along any radial line. Assuming a solution in separated-variable form, show that the full expression for \(\psi\) is $$ \psi(\rho, \phi)=\frac{\rho^{2}}{2} \frac{\sin 2 \phi-2 \phi \cos 2 \alpha}{\sin 2 \alpha-2 \alpha \cos 2 \alpha} $$.

(a) By applying the divergence theorem to the volume integral $$ \int_{V}\left[\phi\left(\nabla^{2}-m^{2}\right) \psi-\psi\left(\nabla^{2}-m^{2}\right) \phi\right] d V $$ obtain a Green's function expression, as the sum of a volume integral and a surface integral, for \(\phi\left(\mathbf{r}^{\prime}\right)\) that satisfies $$ \nabla^{2} \phi-m^{2} \phi=\rho $$ in \(V\) and takes the specified form \(\phi=f\) on \(S\), the boundary of \(V .\) The Green's function \(G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) to be used satisfies $$ \nabla^{2} G-m^{2} G=\delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) $$ and vanishes when \(\mathbf{r}\) is on \(S .\) (b) When \(V\) is all space, \(G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) can be written as \(G(t)=g(t) / t\) where \(t=\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\) and \(g(t)\) is bounded as \(t \rightarrow \infty .\) Find the form of \(G(t)\). (c) Find \(\phi(\mathbf{r})\) in the half space \(x>0\) if \(\rho(\mathbf{r})=\delta\left(\mathbf{r}-\mathbf{r}_{1}\right)\) and \(\phi=0\) both on \(x=0\) and as \(r \rightarrow \infty\) .

(a) Find the form of the solution of Laplace's equation in plane polar coordinates \(\rho, \phi\) that takes the value \(+1\) for \(0<\phi<\pi\) and the value \(-1\) for \(-\pi<\phi<0\) when \(\rho=a\) (b) For a point \((x, y)\) on or inside the circle \(x^{2}+y^{2}=a^{2}\), identify the angles \(\alpha\) and \(\beta\) defined by $$ \alpha=\tan ^{-1} \frac{y}{a+x} \quad \text { and } \quad \beta=\tan ^{-1} \frac{y}{a-x} $$ Show that \(u(x, y)=(2 / \pi)(\alpha+\beta)\) is a solution of Laplace's equation that satisfies the boundary conditions given in (a). (c) Deduce a Fourier series expansion for the function $$ \tan ^{-1} \frac{\sin \phi}{1+\cos \phi}+\tan ^{-1} \frac{\sin \phi}{1-\cos \phi} $$.

Find the Green's function \(G\left(\mathbf{r}, \mathbf{r}_{0}\right)\) in the half-space \(z>0\) for the solution of \(\nabla^{2} \Phi=0\) with \(\Phi\) specified in cylindrical polar coordinates \((\rho, \phi, z)\) on the plane \(z=0\) by $$ \Phi(\rho, \phi, z)= \begin{cases}1 & \text { for } \rho \leq 1 \\ 1 / \rho & \text { for } \rho>1\end{cases} $$ Determine the variation of \(\Phi(0,0, z)\) along the \(z\)-axis.
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