# Calculate Net Charge Contained by the Cube

• MedEx
In summary, the net charge contained by the cube is 4.7849e-10 C. This can be calculated by taking the divergence of the electric field and integrating over the volume of the cube.
MedEx

## 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?

## Homework Equations

net flux= qenc/Eo (Gauss's Law)
Net flux= Ecos(theta)A
A= r^2

## 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.

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

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

#### Attachments

• physics cube a.gif
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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

MedEx said:
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

Can the answer be 4.7849e-10 C ?

coils said:
Can the answer be 4.7849e-10 C ?

Well I thought it could, but now I'm nervous.

MedEx said:
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.

coils said:
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.

MedEx said:
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:

$$\vec{\nabla} . \vec{E} = \rho/\varepsilon_{0}$$
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,
$$\begin{split} Q &= \int dxdydz\ \left(1.65\varepsilon_{0}\right)\\ &= 1.65\varepsilon_{0}\left(2.30\right)^{3}\\ &= 4.7849\times 10^{-10} C \end{split}$$

MedEx and TSny

## 1. How do you calculate the net charge contained by a cube?

To calculate the net charge contained by a cube, you need to know the charge of each individual side of the cube. Then, you can use the formula Q = n x e, where Q is the total charge, n is the number of sides with charge, and e is the charge of each side.

## 2. What is the unit of measurement for net charge?

The unit of measurement for net charge is Coulombs (C).

## 3. Can the net charge of a cube be negative?

Yes, the net charge of a cube can be negative if the charges on each side are not balanced. This means that there are more negative charges than positive charges, resulting in a negative net charge.

## 4. How does the size of the cube affect the net charge?

The size of the cube does not affect the net charge. The net charge is solely determined by the number and distribution of charges on each side of the cube.

## 5. Can you use this formula to calculate the net charge of any 3D shape?

No, this formula specifically applies to cubes. Other 3D shapes may have different methods for calculating net charge, depending on their geometry and distribution of charges.

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