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

Show that an electron in a one-dimensional electri. cal potential \(V(x)=A x^{2},\) where the constant \(A\) is a positive real number, will execute simple harmonic motion about the origin. What is the period of that motion?

A point charge of \(+2.0 \mu C\) is located at \((2.5 \mathrm{~m}, 3.2 \mathrm{~m})\) A second point charge of \(-3.1 \mu \mathrm{C}\) is located at \((-2.1 \mathrm{~m}, 1.0 \mathrm{~m})\). a) What is the electric potential at the origin? b) Along a line passing through both point charges, at what point(s) is (are) the electric potential(s) equal to zero?

A total charge of \(Q=4.2 \cdot 10^{-6} \mathrm{C}\) is placed on a conducting sphere (sphere 1 ) of radius \(R=0.40 \mathrm{~m}\). a) What is the electric potential, \(V_{1},\) at the surface of sphere 1 assuming that the potential infinitely far away from it is zero? (Hint: What is the change in potential if a charge is brought from infinitely far away, where \(V(\infty)=0,\) to the surface of the sphere?) b) A second conducting sphere (sphere 2) of radius \(r=0.10 \mathrm{~m}\) with an initial net charge of zero \((q=0)\) is connected to sphere 1 using a long thin metal wire. How much charge flows from sphere 1 to sphere 2 to bring them into equilibrium? What are the electric fields at the surfaces of the two spheres?

A classroom Van de Graaff generator accumulates a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) on its spherical conductor, which has a radius of \(10.0 \mathrm{~cm}\) and stands on an insulating column. Neglecting the effects of the generator base or any other objects or fields, find the potential at the surface of the sphere. Assume that the potential is zero at infinity.

A charge \(Q=\) \(+5.60 \mu C\) is uniformly distributed on a thin cylindrical plastic shell. The radius, \(R\), of the shell is \(4.50 \mathrm{~cm}\). Calculate the electric potential at the origin of the \(x y\) -coordinate system shown in the figure. Assume that the electric potential is zero at points infinitely far away from the origin.

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