Electrical Field around Spherical Ball at origin

In summary, the problem is to find the electric field along the x-axis for a spherical ball centered at the origin with uniform charge density ##\rho## and radius a. Using the Gauss theorem, the equation for E can be simplified to ##E\times4\pi x^2=\frac{\rho\times4/3\pi a^3}{\epsilon_0}##. To find the electric field, the equation ##E=\frac{\rho}{4\pi\epsilon_0}\int\int\int\frac{\vec{e_{r_{\rho}}}}{r_{\rho}^2}*r^2*sin^2\theta d\phi d\theta dr## is used,
  • #1
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Homework Statement


Spherical Ball centered at origin uniform ##\rho## with a radius a. Find E along x-axis.


Homework Equations


##E = \frac{\rho}{4\pi\epsilon_0}\int\int\int\frac{r^2*sin\theta}{r_\rho^2} d\phi d\theta dr##


The Attempt at a Solution


Evaluate E spherically along the x-axis:
##(x_1, 0, 0): r_\rho^2=(x - x_1)^2 + y^2 + z^2=r^2 + x_1^2 - 2*x_1*cos\theta*cos\phi##

##r_\rho =\sqrt{r^2 + x_1^2 - 2*x_1*cos\theta*cos\phi}##

It seems I am missing a component.
 
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  • #2
Surely you are! ##r_ρ=\sqrt{r^2+x^2_1−2∗r*x_1∗cosθ∗cosϕ}##

But you can solve your problem without any troubles using the Gauss theorem:
$$
E × 4πx^2 = \frac{ρ × 4/3πa^3}{ε_0}
$$
 
  • #3
I need to understand this the long way.

The equation for E I wrote incorrectly
##E = \frac{\rho}{4*\pi*\epsilon_0}*\int\int\int\frac{\vec{e_{r_{\rho}}}}{r_{\rho}^2}*r^2*sin^2\theta d\phi d\theta dr##

How do I find:

## \vec{e_{r_\rho}}##
 
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Related to Electrical Field around Spherical Ball at origin

1. How is the electrical field around a spherical ball at the origin calculated?

The electrical field around a spherical ball at the origin is calculated using the formula E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant, Q is the charge of the spherical ball, and r is the distance from the center of the ball to the point where the electric field is being calculated.

2. What factors affect the strength of the electrical field around a spherical ball at the origin?

The strength of the electrical field around a spherical ball at the origin is affected by the charge of the ball, the distance from the center of the ball, and the material in which the ball is placed (known as the dielectric constant).

3. How does the electrical field around a spherical ball at the origin change with distance?

The electrical field around a spherical ball at the origin follows the inverse square law, which means that as the distance from the center of the ball increases, the strength of the electric field decreases. This relationship is described by the formula E = kQ/r^2.

4. Can the direction of the electrical field around a spherical ball at the origin change?

Yes, the direction of the electrical field around a spherical ball at the origin can change depending on the charge of the ball. If the ball has a positive charge, the electric field will point away from the center of the ball, while a negative charge will cause the electric field to point towards the center of the ball.

5. How does the presence of other charged objects affect the electrical field around a spherical ball at the origin?

If there are other charged objects in the vicinity, their electric fields will also contribute to the overall electric field around the spherical ball at the origin. The resulting electric field can be calculated by adding together the electric fields from each individual object using vector addition.

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