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Use \(V=\frac{k q}{r}, E_{x}=-\frac{\partial V}{\partial x}, E_{y}=-\frac{\partial V}{\partial y},\) and \(E_{z}=-\frac{\partial V}{\partial z}\) to derive the expression for the electric field of a point charge, \(q\)

Short Answer

Expert verified
Question: Derive the expression for the electric field components (Ex, Ey, Ez) of a point charge, q, using the given relations between electric potential (V) and the point charge's properties. Answer: The electric field components (Ex, Ey, Ez) due to a point charge, q, are given by: \(E_x = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}x\) \(E_y = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}y\) \(E_z = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}z\)

Step by step solution

01

Define the potential for point charge

We are given the potential for a point charge as \(V=\frac{kq}{r}\), where k is the electrostatic constant, q is the charge of the point particle, and r is the distance from the point charge.
02

Convert to Cartesian coordinates

For simplicity, we shall first convert the potential to Cartesian coordinates using \(x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi,\) and \(z=r\cos\theta\). After plugging the expression for r in terms of x, y, and z, we have: \(V=\frac{kq}{\sqrt{x^2+y^2+z^2}}\)
03

Calculate partial derivatives for Ex, Ey, and Ez

Now we will differentiate V with respect to x, y, and z to get the electric field components Ex, Ey, and Ez. As V depends on x, y, and z through r, we will apply the chain rule.
04

Find Ex

Using the chain rule, differentiate V with respect to x to get the x-component of the electric field, Ex: \(E_x = -\frac{\partial V}{\partial x} = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}x\)
05

Find Ey

Similarly, differentiate V with respect to y to get the y-component of the electric field, Ey: \(E_y = -\frac{\partial V}{\partial y} = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}y\)
06

Find Ez

Lastly, differentiate V with respect to z to get the z-component of the electric field, Ez: \(E_z = -\frac{\partial V}{\partial z} = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}z\)
07

Summary

We have derived the expressions for the components of the electric field due to a point charge q, using the given formulas and taking partial derivatives. The components Ex, Ey, and Ez are given by: \(E_x = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}x\) \(E_y = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}y\) \(E_z = -\frac{kq}{(x^2+y^2+z^2)^{\frac{3}{2}}}z\) These components can be combined to give the full electric field vector as a function of the position (x, y, z) and the point charge properties.

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Key Concepts

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

Electrostatic Potential
The concept of electrostatic potential, abbreviated as V, is a fundamental aspect in the study of electric fields and forces. It is analogous to the gravitational potential in mechanics, representing potential energy per unit charge. The potential at a point in space is defined as the work done in moving a small test charge from infinity to that point, against the electric field, without any acceleration.

For a point charge, the electrostatic potential is given by the equation:
\[V = \frac{k q}{r}\]
where k is Coulomb's constant (or electrostatic constant), q is the charge of the point particle, and r is the radial distance from the charge. The negative gradient of the electrostatic potential gives us the electric field vector at that point. Understanding the relationship between potential and electric field is crucial in physics since it allows for the computation of forces that charged particles experience.
Partial Derivatives
In mathematics, partial derivatives play an essential role when dealing with functions of multiple variables. They measure how a function changes as only one of the variables is varied, keeping the others fixed.

For a function V(x, y, z), the partial derivatives with respect to x, y, and z are written as &partial;V/&partial;x, &partial;V/&partial;y, and &partial;V/&partial;z. These are the rates of change of V in the direction of each coordinate axis.

In the context of electrostatics, by taking the negative partial derivatives of the electrostatic potential, we obtain the components of the electric field vector. The reason for taking the negative is the convention that the electric field points from higher to lower potential.
Electric Field Components
An electric field is described by a vector field that associates a vector with every point in space around charged objects. However, to fully understand and calculate the influence of this field, we need to break it down into components.

The electric field components Ex, Ey, and Ez are the effects of the electric field along the x, y, and z directions, respectively. These components can be obtained by taking the negative partial derivatives of the electrostatic potential with respect to each coordinate:
\[E_x = -\frac{\partial V}{\partial x}\]
\[E_y = -\frac{\partial V}{\partial y}\]
\[E_z = -\frac{\partial V}{\partial z}\]
These corresponding components are crucial because they allow us to calculate the force a charge would experience from the electric field in a specific direction. Moreover, having the components makes it possible to reconstruct the full vector field by vector addition.
Cartesian Coordinates
The Cartesian coordinate system is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. The location of a point is determined by its distances along the orthogonal x, y, and z axes from a fixed reference point, the origin.

In our exercise, we translated the electrostatic potential equation from a spherical coordinate system, which is naturally suited for a point charge, into a Cartesian coordinate system. We did this to facilitate the calculation of the electric field components. By expressing the potential V in terms of x, y, and z, the calculation becomes a straightforward application of partial derivatives:
\[V = \frac{kq}{\sqrt{x^2 + y^2 + z^2}}\]
This equation reveals the influence of the point charge at every location in space in terms of Cartesian coordinates, which is very useful for computations and visualizations.
Chain Rule in Calculus
The chain rule is a powerful rule in differential calculus for finding the derivative of a composite function. In essence, if a variable z is a function of y, which in turn is a function of x, then z can be considered a function of x as well. The chain rule states that the derivative of z with respect to x is the derivative of z with respect to y multiplied by the derivative of y with respect to x.

In the context of our problem, the electrostatic potential V depends on r, and r depends on x, y, and z. The chain rule allows us to differentiate V with respect to x, y, and z correctly. For instance, when we are finding the component Ex of the electric field, we apply the chain rule as follows:
\[E_x = -\frac{\partial V}{\partial x} = -\frac{\partial V}{\partial r}\cdot\frac{\partial r}{\partial x}\]
By utilizing the chain rule, we can manage the complexity of having an electrostatic potential that depends on a radial distance, which in turn depends on all three Cartesian coordinates.

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Most popular questions from this chapter

One of the greatest physics experiments in history measured the charge-to-mass ratio of an electron, \(q / m .\) If a uniform potential difference is created between two plates, atomized particles - each with an integral amount of charge-can be suspended in space. The assumption is that the particles of unknown mass, \(M,\) contain a net number, \(n\), of electrons of mass \(m\) and charge \(q .\) For a plate separation of \(d,\) what is the potential difference necessary to suspend a particle of mass \(M\) containing \(n\) net electrons? What is the acceleration of the particle if the voltage is cut in half? What is the acceleration of the particle if the voltage is doubled?

A proton is placed midway between points \(A\) and \(B\). The potential at point \(A\) is \(-20 \mathrm{~V}\), and the potential at point \(B\) \(+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

A solid conducting sphere of radius \(R\) is centered about the origin of an \(x y z\) -coordinate system. A total charge \(Q\) is distributed uniformly on the surface of the sphere. Assuming, as usual, that the electric potential is zero at an infinite distance, what is the electric potential at the center of the conducting sphere? a) zero c) \(Q / 2 \pi \epsilon_{0} R\) b) \(Q / \epsilon_{0} R\) d) \(Q / 4 \pi \epsilon_{0} R\)

Which of the following angles between an electric dipole moment and an applied electric field will result in the most stable state? a) \(0 \mathrm{rad}\) d) The electric dipole moment is b) \(\pi / 2\) rad not stable under any condition in c) \(\pi\) rad an applied electric field.

In molecules of gaseous sodium chloride, the chloride ion has one more electron than proton, and the sodium ion has one more proton than electron. These ions are separated by about \(0.24 \mathrm{nm}\). How much work would be required to increase the distance between these ions to \(1.0 \mathrm{~cm} ?\)

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