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What is the excess charge on a conducting sphere of radius \(r=0.15 \mathrm{~m}\) if the potential of the sphere is \(1500 \mathrm{~V}\) and \(V=0\) at infinity?

Short Answer

Expert verified
The excess charge is approximately \( 2.50 \times 10^{-8} \text{ C} \).

Step by step solution

01

Understand the Problem

We have a conducting sphere with a radius of \( r = 0.15 \) m, and a potential \( V = 1500 \) V. We need to find the excess charge on this sphere. The potential at infinity is given as \( V = 0 \).
02

Recall the Formula for Potential of Conducting Sphere

The potential \( V \) on the surface of a sphere is related to its charge \( Q \) and radius \( r \) by the formula: \[ V = \frac{kQ}{r} \]where \( k \) is the electrostatic constant \( 8.99 \times 10^9 \) Nm²/C².
03

Rearrange the Formula to Find Charge

We need to solve for \( Q \). Rearrange the formula to get:\[ Q = \frac{Vr}{k} \]
04

Substitute the Known Values into the Formula

Using the values provided:- \( V = 1500 \) V,- \( r = 0.15 \) m,- \( k = 8.99 \times 10^9 \) Nm²/C²,Substitute these into the equation:\[ Q = \frac{1500 \times 0.15}{8.99 \times 10^9} \]
05

Calculate the Excess Charge

Calculate \( Q \):1. Compute the product \( 1500 \times 0.15 = 225 \).2. Divide by \( 8.99 \times 10^9 \):\[ Q = \frac{225}{8.99 \times 10^9} \approx 2.50 \times 10^{-8} \text{ C} \]

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Key Concepts

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

Potential of Conducting Sphere
The potential of a conducting sphere is an important concept in electrostatics. It refers to the electric potential at the surface of a conducting sphere due to its charge. This potential is uniform over the surface and is given by the formula:
  • \[ V = \frac{kQ}{r} \]
where:
  • \( V \) is the electric potential, measured in volts.
  • \( k \) is the electrostatic constant, \( 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \).
  • \( Q \) is the total charge on the sphere, in coulombs.
  • \( r \) is the radius of the sphere, in meters.
The sphere's potential is determined by the charge and the radius of the sphere, as these affect the distance and distribution of charge at its surface.
This means that the farther you move away from a charged sphere, the lower the potential becomes, approaching zero at infinite distance.
This is often expressed as \( V = 0 \) at infinity, which acts as a reference level of potential.
If you know the potential and radius of the sphere, you can calculate its charge by rearranging the formula to solve for \( Q \).
Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics, describing the interaction between two point charges. It provides the force magnitude between charges and follows this equation:
  • \[ F = k \frac{|q_1q_2|}{r^2} \]
where:
  • \( F \) is the force between two charges, in newtons (N).
  • \( q_1 \) and \( q_2 \) are the amounts of the two charges in coulombs.
  • \( r \) is the separation distance between the charges in meters.
  • \( k \), the electrostatic constant, is \( 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \).
Coulomb's Law tells us that the force grows with increasing charge size but weakens as their distance increases.
It is an inverse-square law, meaning the force is inversely proportional to the square of the distance between charges.
This fundamental understanding helps in explaining the electric potential, as potential relates not just to one point but across a field.
In spheres, these principles combine as each point on a charged sphere contributes to its total potential.
Electrostatic Constant
The electrostatic constant, often designated as \( k \), is crucial in calculations involving electrostatics. It represents the constant of proportionality in both Coulomb's Law and the formula for electric potential.
This constant is valued at \( 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \), ensuring consistency across electrostatic equations, allowing physicists to calculate the electric forces and fields generated by charged particles.
  • It helps quantify the intensity of force between charges separated by a distance \( r \).
  • The constant is applied universally to all electric charge interactions, whether as point charges or distributed on objects like spheres.
The presence of \( k \) in electrostatic equations shows how fundamental forces scale with charge size and distance.
When applied alongside a known radius and charge, it facilitates computing potential, establishing the connectivity of various electrostatic properties.
Through understanding \( k \), students gain a deeper comprehension of electrostatic interactions and how they influence surrounding space fields.

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Most popular questions from this chapter

A particle of charge \(q\) is fixed at point \(P,\) and a second particle of mass \(m\) and the same charge \(q\) is initially held a distance \(r_{1}\) from \(P .\) The second particle is then released. Determine its speed when it is a distance \(r_{2}\) from \(P\). Let \(q=3.1 \mu \mathrm{C}, m=20 \mathrm{mg}\), \(r_{1}=0.90 \mathrm{~mm},\) and \(r_{2}=2.5 \mathrm{~mm}\).

A solid copper sphere whose radius is \(1.0 \mathrm{~cm}\) has a very thin surface coating of nickel. Some of the nickel atoms are radioactive, each atom emitting an electron as it decays. Half of these electrons enter the copper sphere, each depositing \(100 \mathrm{keV}\) of energy there. The other half of the electrons escape, each carrying away a charge \(-e .\) The nickel coating has an activity of \(3.70 \times 10^{8}\) radioactive decays per second. The sphere is hung from a long, nonconducting string and isolated from its surroundings. (a) How long will it take for the potential of the sphere to increase by \(1000 \mathrm{~V} ?\) (b) How long will it take for the temperature of the sphere to increase by \(5.0 \mathrm{~K}\) due to the energy deposited by the electrons? The heat capacity of the sphere is \(14 \mathrm{~J} / \mathrm{K}\).

Identical \(50 \mu \mathrm{C}\) charges are fixed on an \(x\) axis at \(x=\pm 3.0 \mathrm{~m} .\) A particle of charge \(q=-15 \mu \mathrm{C}\) is then released from rest at a point on the positive part of the \(y\) axis. Due to the symmetry of the situation, the particle moves along the \(y\) axis and has kinetic energy \(1.2 \mathrm{~J}\) as it passes through the point \(x=0, y=4.0 \mathrm{~m}\) (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of \(y\) will the particle momentarily stop?

A charge of \(1.50 \times 10^{-8} \mathrm{C}\) lies on an isolated metal sphere of radius \(16.0 \mathrm{~cm} .\) With \(V=0\) at infinity, what is the electric potential at points on the sphere's surface?

An electric field of approximately \(100 \mathrm{~V} / \mathrm{m}\) is often observed near the surface of Earth. If this were the field over the entire surface, what would be the electric potential of a point on the surface? (Set \(V=0\) at infinity.)

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