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1 A charged, conducting sphere of radius \(5.5 \mathrm{~cm}\) sets up a potential of \(75 \mathrm{~V}\) at a radial distance of \(2.2 \mathrm{~m}\) (with \(V=0\) set at infinity). (a) What is the potential on the sphere's surface? (b) What is the surface charge density?

Short Answer

Expert verified
(a) The potential on the surface is higher than 75 V. (b) Surface charge density is calculated using \( \sigma = \frac{Q}{4\pi (0.055)^2} \).

Step by step solution

01

Understanding Potential at Distance

To start, let's recall that the potential due to a charged sphere at a radial distance from its center is given by the formula \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \mathrm{N\,m^2/C^2}) \), \( Q \) is the charge, and \( r \) is the distance from the center. Here, \( V = 75 \, \mathrm{V} \) at a distance \( r = 2.2 \, \mathrm{m} \). We will use this information to calculate the charge \( Q \) on the sphere.
02

Solving for the Charge on the Sphere

Using the equation \( V = \frac{kQ}{r} \), we solve for \( Q \): \[ Q = \frac{Vr}{k} = \frac{75 \, \mathrm{V} \times 2.2 \, \mathrm{m}}{8.99 \times 10^9 \, \mathrm{N\,m^2/C^2}} \] Calculate \( Q \) to find the charge on the sphere.
03

Calculating Potential on Sphere's Surface

The potential on the surface of a conductor is the same as at any point on the surface. For our sphere, consider the surface radius \( r_{\text{surface}} = 0.055 \, \mathrm{m} \). The potential is given as \[ V_{\text{surface}} = \frac{kQ}{r_{\text{surface}}} \]Substitute the charge \( Q \) from Step 2 into this equation to find the potential on the sphere's surface.
04

Calculating Surface Charge Density

Surface charge density \( \sigma \) is defined as the charge per unit area on the surface of the sphere. The formula for surface charge density is \[ \sigma = \frac{Q}{4\pi R^2} \] where \( R \) is the radius of the sphere, which is \( 0.055 \, \mathrm{m} \). Substitute \( Q \) calculated in Step 2 to find \( \sigma \).

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Key Concepts

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

Electric Potential
Electric potential is a measure of the electric potential energy per unit charge at a point in a field. You can think of it as the 'height' of an electrical landscape where a charge would gain or lose energy, much like how water flows downhill. The electric potential due to a point charge is determined using the formula:
  • \[ V = \frac{kQ}{r} \]
where:
  • \( V \) is the electric potential, measured in volts (V).
  • \( k \) is Coulomb’s constant, approximately \( 8.99 \times 10^9 \, \mathrm{N\,m^2/C^2} \).
  • \( Q \) is the charge in coulombs (C).
  • \( r \) is the distance from the charge in meters (m).
The equation shows that the potential decreases as the distance from the charge increases, which is why at infinity, the potential is zero: it is far away from any charges.
In this problem, we start by using the potential at a distance to find the charge on the sphere, which then allows us to calculate the potential at any other point, like on the sphere's surface.
Surface Charge Density
Surface charge density, denoted as \( \sigma \), refers to the amount of electric charge per unit area on the surface of a conductor. It's a crucial concept when dealing with charged conductors because it tells us how the charge is distributed on the surface.
The formula for surface charge density is:
  • \[ \sigma = \frac{Q}{4\pi R^2} \]
Here:
  • \( Q \) is the total charge on the surface of the sphere.
  • \( R \) is the radius of the sphere.
  • The denominator \( 4\pi R^2 \) is the surface area of the sphere.
This measure is useful when we need to comprehend how concentrated a charge is on a particular surface. A high surface charge density indicates a large amount of charge packed into a small area, whereas a low surface charge density shows that the charge is more spread out.
By calculating this value for the charged sphere, we gain insight into how the charge is distributed over the sphere's surface.
Charged Conductor
Conductors are materials that permit free movement of electric charge within them. When a conductor becomes charged, the charges reside solely on its outer surface. This happens because the charges repel each other and are free to move.
For a charged, conducting sphere, the electric field inside is zero because the charges move to the surface and reach a state of equilibrium, distributing themselves uniformly.
  • Think of a conductor as a hollow hill: outside, the landscape varies; inside, it is flat.
  • This property results in all points on the surface having the same potential, meaning the entire surface of the charged sphere is at the same electric potential, calculated using the point charge formula, but with the sphere's radius as the distance.
This knowledge is handy because it allows us to predict and calculate how a sphere or any other conducting object will affect its surroundings and where potential differences reside.
Coulomb's Law
Coulomb's Law is a fundamental principle describing the force between two stationary charged particles. It states that:
  • The force (\( F \)) between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance (\( r \)) between them.
  • Mathematically, it is formulated as:\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where:
  • \( F \) is the electrostatic force between the charges.
  • \( Q_1 \) and \( Q_2 \) are the amounts of the two charges.
  • \( k \) is Coulomb's constant (as previously mentioned).
Though not directly used in the exercise, Coulomb's Law underpins much of electrostatics. It's indispensable for understanding interactions between charges:
  • It explains why oppositely charged objects attract and like charges repel.
  • Helps predict how charges will distribute themselves across surfaces in a uniform manner when forces are balanced (as seen in conductors).
In many ways, it provides the foundation that makes calculations like those in the exercise possible.

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Most popular questions from this chapter

Two tiny metal spheres \(A\) and \(B\), mass \(m_{A}=5.00 \mathrm{~g}\) and \(m_{B}=10.0 \mathrm{~g}\), have equal positive charge \(q=5.00 \mu \mathrm{C}\). The spheres are connected by a massless nonconducting string of length \(d=3.00 \mathrm{~m}\), which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

A spherical drop of water carrying a charge of 30 pC has a potential of \(500 \mathrm{~V}\) at its surface (with \(V=0\) at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop? (c) What is the ratio of the surface charge density on the new drop to that on the original drop?

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 \(b=15 \mathrm{nC} / \mathrm{m}^{2}\) 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?

An electron is projected with an initial speed of \(1.6 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

An infinite nonconducting sheet with a uniform surface charge density sets up parallel equipotential surfaces. Any pair of surfaces differing by \(25.0 \mathrm{~V}\) are separated by \(8.80 \mathrm{~mm}\). (a) What is the magnitude of the surface charge density? (b) If an electron is released near the sheet, does it tend to move from higher to lower potential or vice versa?

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