Question

Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be $$\mathbf{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-\lambda / \lambda} \hat{\mathbf{r}},$$ where $\lambda$ is a new constant of nature (it has dimensions of length, obviously, and is a huge number-say half the radius of the known universe- -so that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds. (a) What is the electric field of a charge distribution $\rho$ (replacing Eq. 2.8 )? (b) Does this electric field admit a scalar potential? Explain briefly how you reached your conclusion. (No formal proof necessary- -just a persuasive argument.) (c) Find the potential of a point charge $q-$ the analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use $\infty$ as your reference point. (d) For a point charge $q$ at the origin, show that $$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} q,$$ where $\mathcal{S}$ is the surface, $\mathcal{V}$ the volume, of any sphere centered at $q$. (e) Show that this result generalizes: $$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} Q_{\mathrm{enc}},$$ for any charge distribution. (This is the next best thing to Gauss's Law, in the new "electrostatics.") (f) Draw the triangle diagram (like Fig. 2.35 ) for this world, putting in all the appropriate formulas. (Think of Poisson's equation as the formula for $\rho$ in terms of $V$, and Gauss's law (differential form) as an equation for $\rho$ in terms of E.)

Name: Problem 49, Chapter 2: Introduction to Electrodynamics
Uploaded: 2022-01-31T18:53:04-08:00
Duration: 13 min 28 s
Channel: Keshav Singh

   Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be
$$\mathbf{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-\lambda / \lambda} \hat{\mathbf{r}},$$
where $\lambda$ is a new constant of nature (it has dimensions of length, obviously, and is a huge number-say half the radius of the known universe- -so that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds.
(a) What is the electric field of a charge distribution $\rho$ (replacing Eq. 2.8 )?
(b) Does this electric field admit a scalar potential? Explain briefly how you reached your conclusion. (No formal proof necessary- -just a persuasive argument.)
(c) Find the potential of a point charge $q-$ the analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use $\infty$ as your reference point.
(d) For a point charge $q$ at the origin, show that
$$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} q,$$
where $\mathcal{S}$ is the surface, $\mathcal{V}$ the volume, of any sphere centered at $q$.
(e) Show that this result generalizes:
$$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} Q_{\mathrm{enc}},$$
for any charge distribution. (This is the next best thing to Gauss's Law, in the new "electrostatics.")
(f) Draw the triangle diagram (like Fig. 2.35 ) for this world, putting in all the appropriate formulas. (Think of Poisson's equation as the formula for $\rho$ in terms of $V$, and Gauss's law (differential form) as an equation for $\rho$ in terms of E.)

Introduction to Electrodynamics

David J. Griffiths,… 3rd Edition

Chapter 2, Problem 49

Instant Answer

Solved by Expert Keshav Singh

01/31/2022

Step 1

We can do this by integrating the force over the charge distribution. The electric field $\mathbf{E}$ is given by $$\mathbf{E}=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho \mathbf{\hat{r}}}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-r / \lambda} d \tau.$$ Show more…

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Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be $$\mathbf{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-\lambda / \lambda} \hat{\mathbf{r}},$$ where $\lambda$ is a new constant of nature (it has dimensions of length, obviously, and is a huge number-say half the radius of the known universe- -so that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds. (a) What is the electric field of a charge distribution $\rho$ (replacing Eq. 2.8 )? (b) Does this electric field admit a scalar potential? Explain briefly how you reached your conclusion. (No formal proof necessary- -just a persuasive argument.) (c) Find the potential of a point charge $q-$ the analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use $\infty$ as your reference point. (d) For a point charge $q$ at the origin, show that $$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} q,$$ where $\mathcal{S}$ is the surface, $\mathcal{V}$ the volume, of any sphere centered at $q$. (e) Show that this result generalizes: $$\oint_{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a}+\frac{1}{\lambda^{2}} \int_{\mathcal{V}} V d \tau=\frac{1}{\epsilon_{0}} Q_{\mathrm{enc}},$$ for any charge distribution. (This is the next best thing to Gauss's Law, in the new "electrostatics.") (f) Draw the triangle diagram (like Fig. 2.35 ) for this world, putting in all the appropriate formulas. (Think of Poisson's equation as the formula for $\rho$ in terms of $V$, and Gauss's law (differential form) as an equation for $\rho$ in terms of E.)

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Key Concepts

Poisson's Equation

Poisson's equation links the scalar potential to the charge density distribution in space. In modified electrostatics, this relationship might include extra terms that emerge from the new force law. Understanding these modifications helps provide a self-consistent field theory that connects potentials, charge distributions, and the electric field.

Triangle Diagram in Electrostatics

This diagram visually represents the interrelationship between the three key formulations in electrostatics: Coulomb's law (which describes the force between point charges), Gauss's law (which connects field flux to enclosed charge), and Poisson's equation (which relates potentials to charge densities). In a modified theory, updating this triangle diagram helps to clarify how changes in one aspect of the theory affect the others and ensures that the overall framework remains consistent.

Gauss's Law and its Modification

Gauss's law relates the net electric flux through a closed surface to the charge enclosed within that surface. When traditional Coulomb's law is amended, Gauss's law may also have corresponding modifications—such as additional volume integrals—to maintain consistency with the new force law. This adjusted form is essential for evaluating the fields of various charge distributions in the modified framework.

Point Charge Potential

This refers to the potential function associated with a point charge, derived by integrating the electric field. In the context of modified electrostatic laws, the point charge potential must be recalculated to reflect any additional terms in the force equation. It forms the basis for understanding interactions at a fundamental level and for validating the modified theory against known limits.

Scalar Potential and Conservative Fields

A scalar potential is a function whose gradient yields the electric field when the field is conservative. The existence of a scalar potential implies that the work done in moving a test charge between any two points is independent of the path taken. Evaluating whether a new form of the electric field admits a scalar potential is crucial for reformulating electrostatics within the framework of potential theory.

Electric Field of a Charge Distribution

This concept deals with determining the electric field resulting from a continuous or discrete distribution of charges. In modified electrostatics, the field is calculated by integrating the contributions from each infinitesimal charge element, taking into account the altered force law. It is essential for formulating a consistent theory that applies to both point charges and extended distributions.

Principle of Superposition

The principle of superposition holds that the total electric field created by a collection of charges is equal to the vector sum of the electric fields produced by each charge individually. This foundational concept remains valid even when Coulomb's law is modified, and it allows for the analysis of complex charge distributions by breaking them down into simpler elements.

Modified Coulomb's Law

This concept concerns alterations to the classical form of Coulomb's law, incorporating additional terms that modify the interaction between charges. Such modifications can include exponential factors or extra distance-dependent corrections that account for new physical insights or experimental discrepancies. Understanding these modifications is key to exploring how fundamental laws of electrostatics might change under new theoretical or experimental conditions.

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Transcript

00:01 For this problem on the topic of electrostatics, we want to imagine that the new force of attraction between two point charges is actually one of a full pi epsilon knot q1, q2 over r squared into 1 plus r over lambda times e to the minus r over lambda r hat.

00:17 Now lambda is a new constant of nature and we have to first find the electric field of a charge distribution row.

00:25 We then want to know if this electric field will admit a scalar potential.

00:30 We then want to find the potential of a point charge q.

00:34 And we want to show that for a point charge q at the origin, that the integral of e .da plus one over a lambda square times the integral of vdor is equal to q of epsilon naught, where s is the surface and v the volume for any sphere centered at q, and then show the generalization of that result.

00:53 And lastly, draw a triangle diagram, putting all the appropriate formulas in.

01:01 So firstly for part a, we have the electric field e equal to 1 over 4 pi epsilon not times the integral of row k hat over k squared into 1 plus k over lambda e to the minus k over lambda detour.

01:30 For part b, we want to know if this electric field would admit a scalar potential, and the answer here is yes, the field of a point charge at the origin is radial and symmetric.

01:53 So the curl of e is still zero, and hence this is also true for any collection of charges.

02:04 For part c, we have the potential v equal to minus the integral from infinity to r of e.

02:21 Dot dl, which is minus 1 over 4 pi epsilon not times q times the integral from infinity to r of 1 over r squared into 1 plus r over lambda, e to the minus r over lambda, dr.

02:47 And this eventually becomes q over 4 pi epsilon not into the integral from r to infinity of 1 over r squared, e to the minus r over lambda d r plus 1 over lambda times the integral from r to infinity of 1 over r e to the minus r over lambda d r now the integral of 1 over r d r.

03:25 Now the integral of 1 over r squared e to the minus r over lambda d r is equal to minus e to the minus r over lambda divided by r minus one over lambda times the integral of minus of e other to the minus r over lambda divided by r d r which will kill the last term and so so the potential v as a function of r is equal to q over 4 pi epsilon not times minus e to the minus r over lambda divided by r.

04:24 And we evaluate this from r to infinity, which gives us the potential q over 4 pi epsilon times e to the minus r.

04:38 Over lambda divided by r.

04:48 For part d, the surface integral of e.

04:55 .da is equal to 1 over 4 pi, epsilon not times q, times 1 over r squared into 1 plus r over lambda, e to the minus r over lambda times 4 pi r squared, which simplifies into q over epsilon knot into 1 plus r over lambda, e to the minus r over lambda.

05:34 And the volume integral over v of v of v detour is equal to q over 4 pi epsilon not times the integral from 0 to r of e to the minus r over lambda over r times r squared times 4 pi d r and this eventually becomes lambda squared q over epsilon knot into minus e to the minus r over lambda into 1 plus r over lambda plus 1.

06:28 And so therefore we have the surface integral of e .da.

06:35 D a plus 1 over lambda squared times the volume integral of v detour equal to q over epsilon not into 1 plus r over lambda e to the minus r over lambda minus 1 plus r over lambda minus 1 plus r over lambda into e to the minus r over lambda into e to the minus r over lambda...

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