# Use Gauss's Law to find the charge contained in the solid hemisphere

• RedDelicious
In summary, the conversation is about a problem from Stewart that was initially posted in the calculus section but is actually closer to a physics problem involving Gauss's Law. The problem involves finding the charge contained in a solid hemisphere using the given electric field. The attempt at a solution involves using a parametric equation for a sphere to model the hemisphere and then using this to parameterize the electric field. However, the solution presented is incorrect due to a mistake in simplifying a term in the middle expression, and the correct answer is found by properly simplifying the term.
RedDelicious

## Homework Statement

edit: I had put this in the calculus section because it was a problem from Stewart but I guess it's closer to a physics problem considering the use of Gauss's Law. My apologies for any confusion this my
[/B]
I've been trying to do this problem without making use of the divergence theorem but have been doing something wrong and cannot figure it out.

Use Gauss’s Law to find the charge contained in the solid hemisphere

$$x^2+y^2+z^2\le a^2,\:z\ge 0$$

if the electric field is

$$E\:=\:<x,\:y,\:2z>$$

## The Attempt at a Solution

[/B]
I use the parametric equation for a sphere to model the hemisphere, with the idea that only the limits of integration over the parameter domain at the end will change, but the parameterization is the same.

$$r\left(\phi ,\theta \right)=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:acos\phi >$$

and using this to parameterize the electric field for the hemisphere

$$F=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:2acos\phi >$$

Using

$$\int \int F\cdot dS\:=\:\int \int F\cdot \left(r_u\:x\:r_v\right)\:dA$$

I get

$$F\cdot \left(r_u\:x\:r_v\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi =a^3sin\phi cos^2\phi$$

$$\int _0^{2\pi }\int _0^{\frac{\pi }{2}}a^3sin\phi cos^2\phi \:d\phi d\theta$$

$$=\frac{2}{3}\pi a^3$$

which is a factor of 4 off, as the answer is $$=\frac{8}{3}\pi a^3\gamma$$

*not including the electric constant

Thanks,
RedDelicious

Last edited:
RedDelicious said:
I get
$$F\cdot \left(r_u\:x\:r_v\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi =a^3sin\phi cos^2\phi$$
This is incorrect. For instance, how will you simplify the first and second term in the middle expression? Moreover, I wonder how come the power of ##a## is 3 not 2.

blue_leaf77 said:
This is incorrect. For instance, how will you simplify the first and second term in the middle expression? Moreover, I wonder how come the power of ##a## is 3 not 2.

Thank you, I wasn't sure if it'd be needed at first, but since I did something wrong here is what I did step by step,

The parametric equation for the sphere is

$$r\left(\phi ,\theta \right)=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:acos\phi >$$

Using this to parameterize the electric field,

$$E=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:2acos\phi >$$
By an example in the book, for a sphere,

$$r_{\phi }\:x\:r_{\theta }\:=\:<a^2sin^2\left(\phi \right)cos\left(\theta \right),\:a^2sin^2\left(\phi \right)sin\left(\theta \right),\:a^2sin\left(\phi \right)cos\left(\phi \right)>$$

and so

$$E\cdot \left(r_{\phi }\:x\:r_{\theta }\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi \\ =a^3sin^3\left(\phi \right)\left[cos^2\left(\theta \right)+sin^2\left(\theta \right)\right]+2a^3sin\left(\phi \right)cos^2\left(\phi \right) \\ =a^3sin^3\left(\phi \right)+2a^3sin\left(\phi \right)cos^2\left(\phi \right) \\ =a^3sin\left(\phi \right)\left(sin^2\left(\phi \right)+2cos^2\left(\phi \right)\right)\\ =a^3sin\left(\phi \right)\left(sin^2\left(\phi \right)+cos^2\left(\phi \right)+cos^2\left(\phi \right)\right)\\ =a^3sin\left(\phi \right)\left(1+cos^2\left(\phi \right)\right)$$

which when substituted into the integral does give me the correct answer of 8a^3pi/3 I repeatedly made a stupid error in simplifying that (1+cos^2(phi)) as just cos^2(phi). Unless you believe that I've erroneously arrived at the answer, I think all is well. Thank you for bringing that step to my attention

## 1. How do I use Gauss's Law to find the charge contained in a solid hemisphere?

To use Gauss's Law, you first need to calculate the electric field at all points on the surface of the hemisphere. This can be done by considering the charge distribution within the hemisphere and using the formula for the electric field of a point charge. Then, you can use Gauss's 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.

## 2. What is the formula for the electric field of a point charge?

The formula for the electric field of a point charge is E = k*q/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the charge of the point charge, and r is the distance from the point charge to the point where the electric field is being calculated.

## 3. How do I calculate the electric flux through a closed surface?

The electric flux through a closed surface is calculated by taking the dot product of the electric field vector and the area vector of each small section of the surface, and then summing up all of these products. Mathematically, it can be represented as Φ = ∫E * dA, where Φ is the electric flux, E is the electric field, and dA is the differential area element.

## 4. What is the permittivity of free space?

The permittivity of free space, denoted by ε0, is a physical constant that represents the ability of a vacuum to permit electric fields. Its value is approximately 8.85x10^-12 C^2/Nm^2.

## 5. Can Gauss's Law be used to find the charge contained in any closed surface?

Yes, Gauss's Law can be used to find the charge contained in any closed surface, as long as the electric field is known at all points on the surface. It is a powerful tool in electromagnetism and is often used to simplify complex charge distributions and calculate the resulting electric fields.

• Calculus and Beyond Homework Help
Replies
3
Views
552
• Calculus and Beyond Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
954
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
871
• Calculus and Beyond Homework Help
Replies
7
Views
2K