/*! 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 47 * A metal sphere with radius \(R... [FREE SOLUTION] | 91Ó°ÊÓ

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* A metal sphere with radius \(R_{1}\) has a charge \(Q_{1} .\) Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Short Answer

Expert verified
(a) \(E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1^2}, V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1}\); (b) Each sphere has \(Q_1 = \frac{Q_{total}R_1}{R_1 + R_2}, Q_2 = \frac{Q_{total}R_2}{R_1 + R_2}\); (c) Equal potential \( \frac{Q_1}{R_1} = \frac{Q_2}{R_2}\); (d) \(E_1\) and \(E_2\) using charges after equilibrium.

Step by step solution

01

Understanding Electrostatics

For a charged sphere, the electric field \(E\) at the surface is given by \(E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}\), where \(Q\) is the charge and \(R\) is the radius. The electric potential \(V\) at the surface is \(V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\), relative to a potential of zero at infinity.
02

Calculating Electric Field and Potential for Sphere 1

Given the charge \(Q_1\) and radius \(R_1\), the electric field at the surface of the first sphere is \(E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1^2}\), and the potential is \(V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1}\).
03

Electrostatic Equilibrium and Charge Redistribution

When the two spheres are connected, charge redistributes to equalize their electric potential. Since they became one system, the total charge \(Q_{total} = Q_1\) is conserved, distributed as \(Q_1 + Q_2 = Q_{total}\) where \(V_1 = V_2\).
04

Calculating Electric Potential for Each Sphere

The potential \(V\) on both spheres is equal at equilibrium: \(\frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2}\). Solve for \(Q_2 = \frac{R_2}{R_1}Q_1\).
05

Finding Total Charge on Each Sphere

Total charge is conserved: \(Q_{total} = Q_1 + Q_2\). Substituting \(Q_2 = \frac{R_2}{R_1}Q_1\), solve for individual charges: \(Q_1 = \frac{Q_{total}R_1}{R_1 + R_2}\) and \(Q_2 = \frac{Q_{total}R_2}{R_1 + R_2}\).
06

Calculating Electric Field at Surface Post-Equilibrium

Using the updated charges, calculate the electric field at the surfaces using \(E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1^2}\) and \(E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2^2}\).

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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. It is a vector field that represents the force experienced by a unit positive charge placed at a point in space. For a charged sphere, like the ones in our exercise, the electric field at the surface is determined by how charge is distributed across the sphere's surface.

When dealing with spheres, the electric field (\(E\)) at the surface is computed using the formula:
  • \(E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}\), where\(Q\) is the charge on the sphere and\(R\) is the radius of the sphere.
This formula tells us that the electric field strength decreases with the square of the radius; larger spheres have weaker fields if the charge is spread out the same way. This relationship is key for understanding how different charged objects interact, especially in a setup like the one where two spheres are connected by a wire.

After the spheres are connected, the fields need reevaluation because the charge distribution changes to achieve electrostatic equilibrium. The field on each sphere is recalculated using their respective new charges, ensuring that both are consistent with the new equilibrium state.
Electric Potential
Electric potential is like a map of potential energy within a field, showing where a charge would "prefer" to go. It's particularly useful, as it simplifies the otherwise complex vector nature of electric fields into scalar quantities.

For a single charged sphere, the electric potential (\(V\)) at the surface is given by:
  • \(V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\), once again,\(Q\) is the charge and\(R\) is the radius of the sphere.
Initially, for our exercise, Sphere 1 has a certain potential based on its charge and size. When the second sphere is introduced and they are connected by a conducting wire, the system evolves to redistribute the charge between the spheres so both have equal potential. This restates one of the fundamental principles of electrostatics: charges move under the influence of electric potential until equilibrium is reached.

What is fascinating here is that even if initially one sphere is uncharged, the connection process ensures both spheres achieve identical potential values, showing the harmony attained among systems via charge redistribution.
Charge Redistribution
Charge redistribution occurs to equalize the electric potential across connected conductive objects. When two conducting objects are connected, charge will flow through the connecting wire until both objects have the same electric potential, achieving electrostatic equilibrium.

In our exercise, we initially have a charged sphere and an uncharged one. Upon connecting them, charge from the originally charged sphere moves to the uncharged sphere until both achieve the same potential. This can be mathematically understood by realizing the equations governing their potentials must equalize:
  • \(\frac{Q_1}{R_1} = \frac{Q_2}{R_2}\)
Here,\(Q_2\) becomes a ratio of radii times the initial charge\(Q_1\), ensuring that the rearranged system remains in equilibrium. The total charge remains conserved, but split according to sphere sizes.

This principle elegantly illustrates how charges naturally align themselves in equilibrium states and informs us about designing systems with desired electric properties. It underscores an important property of conductors in electrostatics: charges will always move to minimize the system's total potential energy while maintaining uniform potential over its surface.

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Most popular questions from this chapter

A positive charge \(q\) is fixed at the point \(x=0, y=0\), and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V=0 ?\) (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a .\) (e) What does the answer to part (b) become when \(x \gg a\) ? Explain why this result is obtained.

Charges \(+q,-q,+q,-q\) areplaced at corners \(A B C D\) of a square of a side \(a\). A charge \(+q\) is placed at its centre. Find the interaction energy of the system

A uniform surface charge of density \(\sigma\) is given to quarter infinite non conducting plane in first quardrant of \(x-y\) plane. Find the \(z\)-component of the electric field at the point \((0,0, z)\). Hence or otherwise find the potential difference between the points \((0,0, d) \&(0,0,2 d)\).

CALC In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y\), where \(A, B\), and \(C\) are positive constants. (a) Calculate the \(x\)-, \(y\)-, and \(z\)-components of the electric field. (b) At which points is the electric field equal to zero?

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$ V(r)=\left\\{\begin{array}{l} \frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a \\ 0 \quad \text { for } r \geq a \end{array}\right. $$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\overrightarrow{\boldsymbol{E}}\) for the regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Eq. (23.23).] Explain why \(\vec{E}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r\). The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(\left.d q=4 \pi r^{2} \rho(r) d r .\right]\) (c) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a\).] Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

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