# Calculating Electric Field of Spherical Charge Distribution

• tome101
In summary, the problem involves computing the electric field generated by a spherically symmetric charged sphere with radius R and charge density of \rho = kr^{2}. This can be done using Gauss' law, which states that the surface integral of E*dA is equal to the charge enclosed divided by the permittivity of free space. By considering a Gaussian sphere of radius r, centered on the sphere of charge, and using symmetry arguments, the surface integral can be simplified to E(4pi*r^2). Integration can also be used to find the charge enclosed by the Gaussian sphere, which is (4pi*k*r^5)/5. By equating these two quantities and solving for E, the electric field can be calculated.
tome101

## Homework Statement

Compute the electric field generated by a spherically symmetric charged sphere of radius R with charge density of $\rho = kr^{2}$

## Homework Equations

$\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$

## The Attempt at a Solution

I know that this question involves the application of Gauss' law but I don't really know how? To be honest I'm a bit sketchy on applying Gauss' law to any question. Any help would really be appreciated.

Imagine a Gaussian sphere of radius r, centered on the actual sphere of charge. What would the surface integral of E*dA be, in terms of r? Remember that due to symmetry, the electric field has to be constant for constant r, and must be entirely radial.

Using integration, can you also find Q_enclosed for this Gaussian sphere?

OK, by integration I've found the charge enclosed by the sphere to be (4pi*k*r^5)/5, but I'm not really sure where to go from here?

From
$\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$
I can see that I need to divide the charge enclosed by epsilon 0 then equate to the surface integral of E*da, but I'm not really sure how to calculate the surface integral of E*da?
Thanks

Ok, I've now been told that the surface integral of E*dA in this case goes to E(4pi*r^2) but I'm still not totally sure why.

Dear student,

Thank you for your question. The solution to this problem involves using Gauss' law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In this case, we can use a spherical Gaussian surface with radius r, centered at the origin, to calculate the electric field at any point outside the charged sphere.

To apply Gauss' law, we first need to determine the charge enclosed by our Gaussian surface. In this case, we can use the given charge density, \rho = kr^{2}, to find the total charge within the sphere. We can do this by integrating the charge density over the volume of the sphere:

Q_{enclosed} = \int_{0}^{R} \rho 4\pi r^2 dr = \frac{4}{5}\pi kR^5

Next, we can calculate the electric flux through the Gaussian surface by taking the dot product of the electric field, \vec{E}, and the differential area element, \vec{dA}. Since the electric field is spherically symmetric, we know that it must be perpendicular to the surface at every point, and thus the dot product simplifies to \vec{E}\cdot \vec{dA} = E dA \cos{\theta}, where \theta is the angle between the electric field and the surface normal.

Using the symmetry of our system, we can see that the electric field must be constant and pointing radially outward at every point on the surface. Therefore, the angle \theta is always 0 and the cosine term is equal to 1. This simplifies our equation to E dA, which is simply the magnitude of the electric field multiplied by the surface area of the Gaussian sphere, 4\pi r^2.

Now, we can set up our equation using Gauss' law:

\oint _S \vec{E} \cdot \vec{dA} = E \int_{0}^{R} 4\pi r^2 dr = \frac{4}{5}\pi kR^5 / \epsilon_0

Solving for the electric field, we get:

E = \frac{kR^3}{5\epsilon_0}

Therefore, the electric field at any point outside the charged sphere is given by this equation. This result makes intuitive sense, as we would expect the electric field

## 1. How do you calculate the electric field of a spherical charge distribution?

To calculate the electric field of a spherical charge distribution, you can use the formula E = k * Q / r^2, where E is the electric field, k is the Coulomb's constant, Q is the total charge of the distribution, and r is the distance from the center of the distribution.

## 2. Can you explain the concept of electric field in relation to a spherical charge distribution?

Electric field is a measure of the force per unit charge exerted by a charged object on other charged objects in its surroundings. In the case of a spherical charge distribution, the electric field is the force per unit charge experienced by a point charge placed at any given distance from the center of the distribution.

## 3. How does the electric field vary with distance from the center of the spherical charge distribution?

The electric field is inversely proportional to the square of the distance from the center of the spherical charge distribution. This means that as the distance increases, the electric field decreases. This relationship is known as the inverse-square law.

## 4. Can you calculate the electric field at any point inside or outside the spherical charge distribution?

Yes, the formula for calculating the electric field of a spherical charge distribution works for any point in space, both inside and outside the distribution. However, the distribution must be symmetric and the point charge must be placed on the same axis as the center of the distribution for the formula to be accurate.

## 5. How does the electric field of a spherical charge distribution compare to that of a point charge?

The electric field of a spherical charge distribution is similar to that of a point charge in the sense that it follows the inverse-square law and is also affected by the total charge of the distribution. However, the electric field of a spherical charge distribution is more complex due to the varying distances from the center of the distribution and the different amounts of charge contained within each spherical shell of the distribution.

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