Evaluating the rate of increase of charge at x=2 yields a value of 16.

[robert25pl]

The law of conservation of charge states that the net current due to flow of charge emanating from a closed surface S is equal to the time rate of decrease of the charge within the volume V bounded by S

$$\oint_{s}J\cdot\,dS=-\frac{d}{dt}\oint_{V}\rho\*d\upsilon$$

Can the time rate of increase of the charge in the cubical volume bounded by x,y,z planes be found?

 

[StatusX]

What do you mean? Of course this applies to any volume, so I don't understand what exactly you're asking.

 

[robert25pl]

when J = -2xi [A/m^2] I have to find the time rate in increase of charge in cubical volume bounded by planes x=0,x=2,y=0,y=2,z=0,z=2. I would like get hint to do this problem. Thanks

 

[StatusX]

Current only passes through the x=0 and x=2 planes, so integrate over these. First find dS, the vector passing out perpendicular to the box with length dA, and the dot this with the current density, and integrate over each surface.

 

[robert25pl]

So I will have?

$$\frac{d}{dt}\oint_{V}\rho\*d\upsilon = -\oint_{S} -2x \cdot 12xdydz\vec{i}$$

$$\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot 12xdydz\vec{i}=96x^{2}$$

Is this correct?

 

[jdavel]

robert,

Your equation says a scalar equals a vector. Is that possible?

 

[robert25pl]

$$\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot 12xdydz\vec{i}=96x^{2}$$

I have solved more difficult problems but I cannot do this one. I'm confused from the beginnig.

 

[StatusX]

Where did that 12x come from? If you drop that and do the integral again, you will get the charging flow in the positive x direction over a 2X2 sheet in a plane parrellel to the x=0 plane. You want to find the total current leaving the box, so you need subtract the current going in at x=0 from the current leaving at x=2.

 

[robert25pl]

$$\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot dydz\vec{i}=8x$$

So at x=2 current would be 16x?

 

[jdavel]

robert said: "I have solved more difficult problems..."

I'm sure you have, because this one is VERY easy; you're making a lot harder than it really is.

In what direction is the current density?

Now define a vector that points out of the cube and in the direction that's perpendicular to the cube face that's at x=0.

What direction is that vector pointing?

If you did this for the other five faces (call them x=2, y=0, y=2...) what direction will each of your vectors be pointing?

 

[robert25pl]

the current density is negative x direction
The vector j points out of the cube and it is in the -y direction

 

[jdavel]

robert,

I have to apologize here. I missed what you said in your post #9. I saw your answer, 16x and thought you were pretty confused. But really you were very close. On the line above (where your integral is) you get dQ/dt = 8x. But then you evaluated that at x=2 and said that it was 16x?

8x evaluated at x=2 isn't 16x. What is it?