# Calculate Net Charge Contained by the Cube

#### MedEx

1. Homework Statement
The figure shows a closed Gaussian surface in the shape of a cube of edge length 3.20 m. It lies in a region where the electric field is given by E=(1.65x+2.86)i + 4.39j + 5.47k N/C, with x in meters. What is the net charge contained by the cube?

2. Homework Equations
net flux= qenc/Eo (Gauss's Law)
Net flux= Ecos(theta)A
A= r^2
3. The Attempt at a Solution
The top, bottom and sides don't matter because they're constant, so I should only need to look at the front and back.
For the front side I had net flux = 2.86(cos (0deg))(10.24)=29.2864
For the back side I had net flux = (1.65*3.2+2.86)(cos 180deg)(10.24) = -83.3536
So the total net flux would be -54.0672.
Multiply this number by 8.55e-12 and I got the net charge to be -4.62275e-10. Any help would be appreciated.

Related Introductory Physics Homework Help News on Phys.org

Homework Helper
Gold Member

#### MedEx

The only figure is the one that is already in the question. Can you not see it?

#### kuruman

Homework Helper
Gold Member
The only figure is the one that is already in the question. Can you not see it?

#### Attachments

• 2.6 KB Views: 147

#### TSny

Homework Helper
Gold Member
What is the value of $x$ on the back face (including sign)?

$\epsilon_0$ is $8.85 \times 10^{-12} \frac{C^2}{N m^2}$, not $8.55 \times 10^{-12} \frac{C^2}{N m^2}$.

#### MedEx

okay so x is heading backwards into the negative direction. so the back face should be (1.65*-3.2+2.86)(cos 180deg)(10.24) = 83.3536
the total net flux is then 83.3536+29.2864 = 112.64
112.64*8.85e-12= 9.96864e-10C for my net charge?
That doesn't seem like it works either

#### TSny

Homework Helper
Gold Member
the back face should be (1.65*-3.2+2.86)(cos 180deg)(10.24) = 83.3536
This looks set up OK, but I get a different numerical result here.

#### MedEx

Did you get 24.781? I might have just took the original 83 number and flipped the sign.
alright 24.7808+29.2864=54.0672
54.0672*8.85e-12 is 4.78495e-10C

#### coils

Can the answer be 4.7849e-10 C ?

#### MedEx

Can the answer be 4.7849e-10 C ?
Well I thought it could, but now I'm nervous.

#### coils

Well I thought it could, but now I'm nervous.
Well, my approach to the solution was just take the divergence of the electric field and integrate over the volume.

#### MedEx

Well, my approach to the solution was just take the divergence of the electric field and integrate over the volume.
What do you get when you try to do it that way? I'm really not very good at calc so that wouldn't be my go to.

#### TSny

Homework Helper
Gold Member
Did you get 24.781? I might have just took the original 83 number and flipped the sign.
alright 24.7808+29.2864=54.0672
54.0672*8.85e-12 is 4.78495e-10C
This looks good to me. @coils method is nice, if you are familiar with the divergence equation for the electric field.

You might consider how many significant figures you should keep in your answer.

#### MedEx

Yup that worked. Thanks to everyone c:

#### coils

$$\vec{\nabla} . \vec{E} = \rho/\varepsilon_{0}$$
Now for the calculation of the LHS, we have;
$$\begin{split}\vec{\nabla} .\vec{E} &= \left(\frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\right) . \left(\left(1.65x + 2.86\right)\hat{x} + 4.39\hat{y} + 5.47\hat{z}\right)\\ &= \frac{\partial \left(1.65x + 2.86\right)}{\partial x} + \frac{\partial \left(4.39\right)}{\partial y} + \frac{\partial \left(5.47\right)}{\partial z}\\ &= 1.65 \end{split}$$

Multiply with the $\varepsilon_{0}$ and integrate over the volume of the cube as,
$$\begin{split} Q &= \int dxdydz\ \left(1.65\varepsilon_{0}\right)\\ &= 1.65\varepsilon_{0}\left(2.30\right)^{3}\\ &= 4.7849\times 10^{-10} C \end{split}$$

"Calculate Net Charge Contained by the Cube"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving