/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 A charge \(q\) is distributed un... [FREE SOLUTION] | 91Ó°ÊÓ

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A charge \(q\) is distributed uniformly throughout a spherical volume of radius \(R .\) Let \(V=0\) at infinity. What are (a) \(V\) at radial distance \(r

Short Answer

Expert verified
(a) \( V(r) = \frac{q}{8 \pi \varepsilon_0 R^3}(3R^2 - r^2) \). (b) Potential difference is \(-\frac{q}{4 \pi \varepsilon_0 R}\).

Step by step solution

01

Understanding the Problem

We need to find the electric potential \( V \) at a point inside a sphere for \( r < R \), and then find the potential difference between the points at \( r = R \) and \( r = 0 \). The charge \( q \) is uniformly distributed within a sphere of radius \( R \).
02

Calculate Electric Field Inside the Sphere

For a charge uniformly distributed in a sphere, the electric field inside the sphere at a distance \( r \) from the center (where \( r < R \)) can be derived from Gauss’s law. It is given by:\[E(r) = \frac{q r}{4 \pi \varepsilon_0 R^3}\]This indicates that the electric field depends linearly on \( r \).
03

Calculate Potential Inside the Sphere

The potential \( V \) at a distance \( r < R \) inside the sphere, relative to the potential at infinity \((V = 0)\), is given by integrating the electric field from infinity to the point \( r \):\[V(r) = - \int_{\infty}^{r} E(r') \, dr' = \frac{q}{8 \pi \varepsilon_0 R^3} (3R^2 - r^2)\]Here, the integration is done considering that the electric field outside the sphere is zero.
04

Find Potential Difference Between r = R and r = 0

We need to find the potential difference between \( r = R \) and \( r = 0 \):\[\Delta V = V(R) - V(0)\]From Step 3, we substitute the values: \[V(R) = \frac{q}{8 \pi \varepsilon_0 R}\] and \[V(0) = \frac{3q}{8 \pi \varepsilon_0 R}\]Therefore, \[\Delta V = \left(\frac{q}{8 \pi \varepsilon_0 R}\right) - \left(\frac{3q}{8 \pi \varepsilon_0 R}\right) = -\frac{2q}{8 \pi \varepsilon_0 R} = -\frac{q}{4 \pi \varepsilon_0 R}\]
05

Conclusion

The potential \( V \) at a radial distance \( r < R \) is \( V(r) = \frac{q}{8 \pi \varepsilon_0 R^3} (3R^2 - r^2) \). The potential difference between \( r = R \) and \( r = 0 \) is \(-\frac{q}{4 \pi \varepsilon_0 R}\).

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Key Concepts

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

Electric Field
The electric field is a crucial concept in electrostatics, representing the force per unit charge that would act on a small positive test charge in the vicinity of other charges. In this context, we have a uniformly distributed charge within a sphere. Using Gauss's Law, we can find the electric field inside the sphere.

The law helps us understand how charges behave within symmetrical configurations, like this sphere. For a distance from the center of the sphere where the radius is lesser than the sphere's radius, the electric field is given by:
  • \( E(r) = \frac{q r}{4 \pi \varepsilon_0 R^3} \)
This indicates that the electric field inside the sphere increases linearly with the distance from the center, directly proportional to the radius \( r \).

The proportionality to \( r \) means that if you move further from the center inside this sphere, the electric field strength increases until you reach the surface of the sphere.
Potential Difference
Potential difference is related to the work done in moving a charge from one point to another within an electric field. In this problem, we are interested in the electric potential inside a spherical charge distribution.

Electric potential \( V \) at a distance \( r \) inside a charged sphere can be calculated by integrating the electric field from a point at infinity to \( r \). This integral accounts for how the field changes as one moves inside the sphere:
  • \( V(r) = \frac{q}{8 \pi \varepsilon_0 R^3} (3R^2 - r^2) \)
This expression shows that within the sphere, the potential decreases as we move from the center \( r = 0 \) to the edge \( r = R \).

Moreover, the potential difference between the edge and the center of the sphere is:
  • \( \Delta V = V(R) - V(0) = -\frac{q}{4 \pi \varepsilon_0 R} \)
This negative sign indicates a loss in potential energy as we move towards the center of the sphere from its surface.
Gauss's Law
Gauss's Law is a pivotal principle in electrostatics, connecting the electric fields at a boundary with the charges within. It simplifies the calculation of the electric field when dealing with symmetric charge distributions, such as spheres.

In this context, Gauss's Law allows the calculation of electric fields inside the sphere by considering a virtual Gaussian surface within. The formula used arises from Gauss's Law and symmetry. It implies:
  • The electric field inside a uniformly charged sphere varies linearly with the radius.
  • Outside such a sphere, if the sphere is uniformly charged, the electric field is akin to that of a point charge located at the center.
Understanding how Gauss’s Law works is key because it helps reduce complex field calculations into manageable equations by leveraging the symmetry properties of the charge distribution. By applying Gauss’s Law, we gain a better grasp of how charges create electric fields in symmetric configurations.

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Most popular questions from this chapter

A nonuniform linear charge distribution given by \(\lambda=\) \(b x\), where \(b\) is a constant, is located along an \(x\) axis from \(x=0\) to \(x=0.20 \mathrm{~m}\). If \(b=20 \mathrm{nC} / \mathrm{m}^{2}\) and \(V=0\) at infinity, what is the electric potential at (a) the origin and (b) the point \(y=0.15 \mathrm{~m}\) on the \(y\) axis?

(a) If Earth had a uniform surface charge density of \(1.0\) electron \(/ \mathrm{m}^{2}\) (a very artificial assumption), what would its potential be? (Set \(V=0\) at infinity.) What would be the (b) magnitude and (c) direction (radially inward or outward) of the electric field due to Earth just outside its surface?

Two uniformly charged, infinite, nonconducting planes are parallel to a \(y z\) plane and positioned at \(x=-50 \mathrm{~cm}\) and \(x=+50\) \(\mathrm{cm} .\) The charge densities on the planes are \(-50 \mathrm{nC} / \mathrm{m}^{2}\) and \(+25\) \(\mathrm{nC} / \mathrm{m}^{2}\), respectively. What is the magnitude of the potential difference between the origin and the point on the \(x\) axis at \(x=+80 \mathrm{~cm} ?\) (Hint: Use Gauss' law.)

Two large, parallel, conducting plates are \(12 \mathrm{~cm}\) apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of \(3.9 \times 10^{-15} \mathrm{~N}\) acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates?

Suppose that in a lightning flash the potential difference between a cloud and the ground is \(1.0 \times 10^{9} \mathrm{~V}\) and the quantity of charge transferred is \(30 \mathrm{C}\). (a) What is the change in energy of that transferred charge? (b) If all the energy released could be used to accelerate a \(1000 \mathrm{~kg}\) car from rest, what would be its final speed?

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