What is the method for determining charge enclosed in a cube using Gauss's law?

[Reshma]

I have the electric field in a region give by:
\vec E = \frac{k}{\epsilon_0 a^4} \left(ax^2\hat x + yz^2\hat y + y^2z\hat z\right)
where 'k' and 'a' are constants.
There are few questions I need to solve.

1: Is the field conservative?
A: Yes. I computed the curl and found it equal to zero.
\nabla \times \vec E = 0

2: Calculate the charge density at a point P(x,y,z).
A: I applied the differential form of Gauss's law.
\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}
\rho is the charge density.
So, I got the charge density as:
\rho = \epsilon_0 \left[2ax + y^2 + z^2\right]
Please verify if my method is correct.

3: Determine the charge enclosed in a cube of side 'a' with one of its corners at the origin and sides parallel to x,y and z axes.
A: I applied Gauss's law here too!
\int \vec E \cdot d\vec a = \frac{Q_{enclosed}}{\epsilon_0}
How do I calculate the flux here?

 

[quantumdude]

2: Calculate the charge density at a point P(x,y,z).
A: I applied the differential form of Gauss's law.
\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}
\rho is the charge density.
So, I got the charge density as:
\rho = \epsilon_0 \left[2ax + y^2 + z^2\right]
Please verify if my method is correct.

Your method is correct.

3: Determine the charge enclosed in a cube of side 'a' with one of its corners at the origin and sides parallel to x,y and z axes.
A: I applied Gauss's law here too!

Right again.

\int \vec E \cdot d\vec a = \frac{Q_{enclosed}}{\epsilon_0}
How do I calculate the flux here?

Calculate it over each face of the cube, one at a time. Here's a hint: For the two faces that are parallel to the yz-plane, the unit normals are \hat{i} and -\hat{i}.

 

[Tide]

Problem 3 is ambiguous since anyone of the 8 corners of the cube could be at the coordinate origin - which is probably a hint at what the answer will be! :)

All you need to do is recognize that d\vec a is in the direction of the normal to the surface and consists of products of dxdy, dxdz and dydz - then integrate over all 6 faces of the cube.

 

[Reshma]

Thanks Tom Mattson and Tide for looking into my problem!

I have the electric field in a region give by:
\vec E = \frac{k}{\epsilon_0 a^4} \left(ax^2\hat x + yz^2\hat y + y^2z\hat z\right)
where 'k' and 'a' are constants.

2: Calculate the charge density at a point P(x,y,z).
A: I applied the differential form of Gauss's law.
\vec \nabla \cdot \vec E = \frac{\rho}{\epsilon_0}
\rho is the charge density.
So, I got the charge density as:
\rho = \epsilon_0 \left[2ax + y^2 + z^2\right]

Er..actually I made a serious blunder here .
I forgot to multiply the constant factor in the electric field expression. So the charge density is:
\rho = \frac{k}{\epsilon_0 a^4} \epsilon_0 \left[2ax + y^2 + z^2\right]
\rho = \frac{k}{a^4} \left[2ax + y^2 + z^2\right]

 

[Reshma]

Problem 3 is ambiguous since anyone of the 8 corners of the cube could be at the coordinate origin - which is probably a hint at what the answer will be! :)
All you need to do is recognize that d\vec a is in the direction of the normal to the surface and consists of products of dxdy, dxdz and dydz - then integrate over all 6 faces of the cube.

Thanks, I tried it. There are 6 faces in all.
1] x = a, d\vec a = dydz\hat x

\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a a^2 dydz = \frac{ak}{\epsilon_0}

2] x = 0, d\vec a = -dydz\hat x
Intergal vanishes here!

3] y = a, d\vec a = dxdz\hat y

\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a az^2 dxdz = \frac{ak}{3\epsilon_0}

4] y = 0, d\vec a = -dxdz\hat y
Integral vanishes here too.

5] z = a, d\vec a = dxdy\hat z

\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a ay^2 dxdz = \frac{ak}{3\epsilon_0}

6] z = 0, d\vec a = -dxdy\hat z
Integral vanishes here.

So the total flux is:
\frac{ak}{\epsilon_0} + \frac{ak}{3\epsilon_0} + \frac{ak}{3\epsilon_0} = \frac{5ak}{3\epsilon_0}

Hence, total charge enclosed is:
Q_{enclosed} = \epsilon_0 \frac{5ak}{3\epsilon_0} = \frac{5ak}{3}

 

[Reshma]

An alternative method

I discovered yesterday that there is yet another alternative method to solve this problem. I can use the Green's theorem over a volume element.
Q_{enclosed} = \int_{all space} \rho dV
dV is the volume element given by dV = dxdydz
So,
Q_{enclosed} = \frac{k}{a^4}\int_0^a \int_0^a \int_0^a \left[2ax + y^2 + z^2\right]dx dy dz
The evaluation of this integral gives me the same result as above.
Q_{enclosed} = \frac{5ak}{3}
So my answer is correct!

 

[Tide]

Way to go, Reshma!