# Electric Field - 1/8 of Sphere

• luiseduardo
In summary, the electric field created by 1/8 of a sphere charged with density of charge equal σ is σ/4.
luiseduardo

## Homework Statement

Calculate the intensity of the electric field (in the origin of cartesian axes) created by 1/8 of a certain sphere charged with density of charge equal σ.

$$\frac{{\sigma \sqrt 3 }}{{16{\varepsilon _0}}}$$

E = KQ/R²
Flux = Q/ε
E.A = Q/ε

## The Attempt at a Solution

Well, I tried to calculate the electric flux in 1/8 of the sphere (see figure). Then, knowing the area of 1/8 sphere, we could calculate the electric field using gauss law. But, after all, the answer of the problem is different.

You'll have to explain your process more -- what values you used, and how you fit those into your equations -- since I don't quite understand it. What flux is going through the sphere?

The flux in total sphere is x, so flux now is x/8.. No?

Flux from what, specifically? That's true if there's a point charge at the center of the sphere, but not the case if there's only an 1/8 of the sphere. I don't think Gauss's law is going to be very helpful.

There's really no symmetry to exploit, so I think this is just going to be a full out vector integration. It will look something like this:

$\int^{R}_{0}\int^{\frac{\pi}{2}}_{0}\int^{\frac{\pi}{4}}_{0}d\stackrel{\rightarrow}{E}dV$

Where $dV = \rho^{2}sin\theta d\rho d\phi d\theta$, and $\stackrel{\rightarrow}{r} = \left\langle \rho sin\phi cos\theta , \rho sin\phi sin\theta , \rho cos\theta \right\rangle$

But, this problem is from a middle school book, so don't think spherical coordinate is the unique solution. Did anyone knows other way to solve this problem ?

Using spherical coordinates, the potential at the origin seemingly must be kσA/r since all charge is r away from the observation point. A = area of 1/4 of hemisphere, k = 1/4πε0.

Computing the E field as - del V seemingly gives |E| = kσπ/2

Need some expert help here ...

EDIT: I realize my computations are overly simple - the gradient at the origin must be evaluated with excursions from the origin in all three coordinates - and when you do that you get dV a function not only of dr but of r, dθ and dø as well. Not that that helped me solve the problem.

mfb, tsny et al ... ?

Last edited:
Using cartesian coordinates?

luiseduardo said:
Using cartesian coordinates?
The integrations using cartesian would be a lot messier. This problem is a natural for spherical.

Hum... If we draw a cube over the sphere of picture? We can calculated something?

There is some symmetry that you can exploit. For 1/8 of a sphere occupying one octant of the xyz coordinate system, the net E vector at the origin will have equal x, y, z components. So, you just need to evaluate one component, say the z-component.

TSny said:
There is some symmetry that you can exploit. For 1/8 of a sphere occupying one octant of the xyz coordinate system, the net E vector at the origin will have equal x, y, z components. So, you just need to evaluate one component, say the z-component.

Thanks, but could you be more specific? Because how to calculate this component without using spherical coordinates.. I'm confused..

I was just saying that if you are going to integrate to find E, then you can just do the integration for the z-component since the other components will be the same. Using spherical coordinates, the integral is not difficult.

However, if you want to avoid integration altogether, then you can resort to subterfuge.

Some hints:

(1) First consider finding the electric field at the center of a uniformly charged hemisphere. You should be able to relate the net E field at the origin of the hemisphere to the z-component of the E field of the 1/8 sphere.

(2) To handle the hemisphere, imagine putting a point charge q at the origin. The net force of the hemisphere on q equals qE where E is the net field at the origin due to the hemisphere. So, if you can get the force on q you can get E at the center of the hemisphere.

(3) The force that the hemisphere exerts on q equals the force that q exerts on the hemisphere.

(4) See if you can get the net force of q on the hemisphere without any detailed integration.

1 person
Thank you very much TSny ! I found the electric field created by hemisphere is σ/4.e0, so, my problem is 1/4 of a hemisphere, so, axis z = y = x = σ/16.e0, now, using pitagoras, we can find the answer !
:)

luiseduardo said:
Thank you very much TSny ! I found the electric field created by hemisphere is σ/4.e0, so, my problem is 1/4 of a hemisphere, so, axis z = y = x = σ/16.e0, now, using pitagoras, we can find the answer !
:)

Sounds good!

## 1. What is an electric field?

An electric field is a physical quantity that describes the influence an electric charge will have on other charges in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is an electric field created?

An electric field is created by a source charge, such as an electron or a proton. The electric field produced by the source charge will exert a force on any other charges in its vicinity.

## 3. What is the formula for calculating the electric field of 1/8 of a sphere?

The formula for calculating the electric field of 1/8 of a sphere is E = (1/4πε0) * (Q/r2), where E is the electric field, Q is the charge of the sphere, r is the distance from the center of the sphere, and ε0 is the permittivity of free space.

## 4. How does the electric field change as you move farther away from the 1/8 sphere?

The electric field decreases as you move farther away from the 1/8 sphere. This is because the electric field is inversely proportional to the square of the distance from the source charge.

## 5. Can the electric field of 1/8 of a sphere be negative?

Yes, the electric field of 1/8 of a sphere can be negative. This would occur if the charge of the sphere is negative, as the direction of the electric field is determined by the direction of the force exerted on a positive test charge.

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