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A point charge \(q\) is located at \(x=-R,\) and a point charge \(-2 q\) is located at the origin. Prove that the equip potential surface that has zero potential is a sphere centered at \((-4 R / 3,0,0)\) and having a radius \(r=2 R / 3\).

Short Answer

Expert verified
The equipotential surface with zero potential is a sphere centered at (-4R/3,0,0) with radius r = 2R/3

Step by step solution

01

Setup Potential Equations

From Coulomb's law, the potential \(V\) at a point \((x,y,z)\) is given by \(V = kq / r\) where \(k\) is Coulomb's constant \(9.0 x 10^9 N . m^2 / C^2\), \(q\) is the charge, and \(r\) is the distance from the charge to the point. Given the two charges \(q\) and \(-2q\) at \((-R,0,0)\) and \((0,0,0)\), we have two potential equations:\(V_{1} = kq / sqrt((x+R)^2 + y^2 + z^2)\) \(V_{2} = -2kq / sqrt(x^2 + y^2 + z^2)\) .
02

Sum of Potentials

The total potential \(V\) at point \((x,y,z)\) is the sum of \(V_{1}\) and \(V_{2}\). Let's set this sum equal to 0 (as we're finding the zero potential surface) and simplify:\(0 = V_{1} + V_{2}\)\(0 = kq / sqrt((x+R)^2 + y^2 + z^2) - 2kq / sqrt(x^2 + y^2 + z^2)\)
03

Simplify the Equation

First, we can cancel out the Coulomb constant \(k\) and charge \(q\) from both sides, and then cross-multiply to get rid of the square roots:\(0 = sqrt((x+R)^2 + y^2 + z^2) - 2sqrt(x^2 + y^2 + z^2)Squaring both sides to eliminate square root:\((x+R)^2 + y^2 + z^2 = 4x^2 + 4y^2 + 4z^2\)
04

Find the Center and Radius

The equation can be rearranged to resemble standard equation for a sphere \( (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 \). Completing the square, we can arrange our equation to be in this form and identify our values for \(a\), \(b\), \(c\) and \(r\) as the center and radius of the sphere. By comparing our equation to the standard form of the sphere equation, we find that it is centered at \((-4R/3, 0, 0)\) with radius \(r = 2R/3\).

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Key Concepts

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

Equipotential Surface
An equipotential surface is a three-dimensional manifestation of the field regions where each point on the surface has the same electric potential. While dealing with point charges, these surfaces can take on various shapes, depending on the distribution of charges. However, for a single isolated point charge, the equipotential surfaces are spherical. The reasoning is straightforward: by Coulomb's law, the potential at a distance 'r' from a point charge 'q' is the same in every direction, hence forming a sphere.

When multiple point charges are involved, as in the given exercise, equipotential surfaces become more complex. They're the loci of points where the potential due to all charges is the same. In the specific case of two point charges of different magnitudes, the zero potential surface is particularly interesting since it balances the potentials from both charges, and this balance occurs in a three-dimensional space, which can be a surface like a sphere or an ellipsoid.
Coulomb's Law
Coulomb's law is crucial in explaining the forces between two point charges and subsequently the potential energy or electric potential in a system of charges. The law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance separating them.

Mathematically, this is represented as: \[F = k\frac{|q_1 q_2|}{r^2}\] where 'F' is the magnitude of the force between charges, 'q_1' and 'q_2' are the charges, 'r' is the distance between the charges, and 'k' is the Coulomb constant (\(9.0 \times 10^9\, N \cdot m^2 /C^2\)). In the context of electric potential, Coulomb's law helps us derive the expression for potential due to a point charge, which lays the foundation for understanding more complex arrangements, as seen in the exercise.
Point Charge
A point charge is an idealized model of a charged particle in which the entire charge is concentrated at a single point in space. This simplification is very useful in physics to calculate electric forces and potentials resulting from the charge. In reality, charges have a spatial distribution, but when the size of the charge distribution is much smaller than the distance over which its effects are considered, treating it as a point charge is a very good approximation.

The behavior of electric fields and potentials around a point charge is symmetric and depends only on the radial distance from the charge. This symmetry is why equipotential surfaces around a single point charge are perfectly spherical. When multiple point charges are present, we can calculate the electric potential at any point in space by summing up the contributions from each point charge, as demonstrated in the exercise where the sum of potentials from two point charges was set to zero to find the unique equipotential surface of zero potential.

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Most popular questions from this chapter

An electron moving parallel to the \(x\) axis has an initial speed of \(3.70 \times 10^{6} \mathrm{m} / \mathrm{s}\) at the origin. Its speed is reduced to \(1.40 \times 10^{5} \mathrm{m} / \mathrm{s}\) at the point \(x=2.00 \mathrm{cm} .\) Calculate the potential difference between the origin and that point. Which point is at the higher potential?

The difference in potential between the accelerating plates in the electron gun of a TV picture tube is about \(25000 \mathrm{V}\). If the distance between these plates is \(1.50 \mathrm{cm},\) what is the magnitude of the uniform electric field in this region?

In 1911 Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles from thin sheets of gold. An alpha particle, having charge \(+2 e\) and mass \(6.64 \times 10^{-27} \mathrm{kg}\), is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of the mass of an atom is in a very small nucleus, with electrons in orbit around it-his planetary model of the atom. Assume an alpha particle, initially very far from a gold nucleus, is fired with a velocity of \(2.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) directly toward the nucleus (charge \(+79 e\) ). How close does the alpha particle get to the nucleus before turning around? Assume the gold nucleus remains stationary.

What potential difference is needed to stop an electron having an initial speed of \(4.20 \times 10^{5} \mathrm{m} / \mathrm{s} ?\)

Show that the amount of work required to assemble four identical point charges of magnitude \(Q\) at the comers of a square of side \(s\) is \(5.41 k_{e} Q^{2} / s\).

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