Electricity and Magnetism Question

[craig.16]

Homework Statement

The potential caused by an array of charges in a region of space can be described by the equation V(x,y,z)=2x+xy+y²-xz² volts, where all distances are measured in metres. If a charge of 2 mC is moved from the position r₁=(1,2,0)m to r₂=(3,1,0), calculate the resulting change in the potential energy of the system.

Calculate the electric field associated with this potential.

Determine the force that acts on the charge when it is at position r₁.

Homework Equations

E=V/d
F=E/q
V=k*(q/r)

The Attempt at a Solution

For the first part I did:
(delta)r=(3,1,0)-(1,2,0)=(2,-1,0)
|(delta)r)=SQRT(5)
V=(8.99*10⁹)(2*10^-3/SQRT(5))=8.04*10⁶J
is this correct?

Im not so confident on the above answer as the electric potential energy involves two point charges but this only includes one. Do i just square this charge to get the right answer? Also I think potential energy is U not V, any clarification on this would be great.

For the second part I did:
inputting (delta)r=(2,-1,0)m into V(x,y,z) i get:
V=2(2)+(2)(-1)+(-1)²-(2)(0)²=4-2+1=3V
Using E=V/d I get:
E=3/SQRT(5)=1.341641V/m (6dp)
Is this correct?

For the last part I did:
F=E/q=1.341641/(2*10^-3)=671N
Is this correct?

 

[collinsmark]

For the first part I did:
(delta)r=(3,1,0)-(1,2,0)=(2,-1,0)
|(delta)r)=SQRT(5)
V=(8.99*10⁹)(2*10^-3/SQRT(5))=8.04*10⁶J
is this correct?

No, that approach isn't going to work for this problem.

First, plug r₁=(1,2,0)m into your V(x,y,z)=2x+xy+y²-xz² volts find V₁. Do the same with r₂ to find V₂. The difference in potential is V₂ - V₁.

Im not so confident on the above answer as the electric potential energy involves two point charges but this only includes one.

Actually, there are many charges involved for this problem, as stated in the problem statement, "...potential caused by an array of charges..."

Do i just square this charge to get the right answer?

Square the charge? :uhh: No, no. Don't do anything like that. :tongue2: Once you find the electrical potential at some point in space, you can find the potential energy of a particular charge at that point in space by multiplying the potential by that particular charge.

Similarly, as in you could do for this problem, you can find the difference in electrical potential of two points in space, and multiply that potential difference by a particle's charge to find the difference in potential energy of that particular charge, between those two points in space.

Also I think potential energy is U not V, any clarification on this would be great.

For a particular charge q, and considering only the electrical potential energy and no other types of potential energy,

ΔU = qΔV.​

Here "Δ" is "delta;" same thing as "difference."

For the second part I did:
inputting (delta)r=(2,-1,0)m into V(x,y,z) i get:
V=2(2)+(2)(-1)+(-1)²-(2)(0)²=4-2+1=3V
Using E=V/d I get:
E=3/SQRT(5)=1.341641V/m (6dp)
Is this correct?

No. The E = V/d only works if E is constant over all space (or at least constant over all space for the region being worked with). And, the equation only applies if d is in a parallel direction to E.

But here E is not constant, and we don't even know what direction E is in yet (or even if it has a constant direction, which it it turns out hasn't), so that equation doesn't apply. You need to use

$$\vec E = -\nabla V$$

where [itex] \nabla [/tex] is the "del" operator (not the same thing as Δ).

$$\nabla = \frac{\partial}{\partial x} \hat x + \frac{\partial}{\partial y} \hat y + \frac{\partial}{\partial z} \hat z$$

[Edit: also note that you shouldn't be plugging in any particular position values for the second part. The answer to the second part should be true for all r.]

For the last part I did:
F=E/q=1.341641/(2*10^-3)=671N
Is this correct?

The force is not the electric field divided by q. Rather the force is the electric field multiplied by q! (That, and you'll still need to re-do your calculation of E.)

 

[craig.16]

Thanks for the help, the amount of stuff I got wrong here is woefully bad haha. I recall the equation $$\Delta$$U=q$$\Delta$$V now except I used to know it as $$\Delta$$W=q$$\Delta$$V from A levels and that's what probably threw me off since I am now used to doing potential energy as $$\Delta$$U not $$\Delta$$W. Anyways, below is my updated answers:

Part 1:
V₁=2(1)+(1)(2)+(2)²-(1)(0)²
=2+2+4
=8V

V₂=2(3)+(3)(1)+(1)²-(3)(0)²
=6+3+1
=10V

$$\Delta$$V=2V

$$\Delta$$U=(2*10^-3)(2)=4mJ

For Part 2 I don't know what to input for x, y and z to get -$$\nabla$$V since its for all values of r not the ones given.

Also how do you get your letters to adjust to the same size as the symbols, I am new to these forums so I am not quite sure how to do these sort of things yet.

 

[collinsmark]

Part 1:
V₁=2(1)+(1)(2)+(2)²-(1)(0)²
=2+2+4
=8V

V₂=2(3)+(3)(1)+(1)²-(3)(0)²
=6+3+1
=10V

$$\Delta$$V=2V

$$\Delta$$U=(2*10^-3)(2)=4mJ

'Looks good to me!

For Part 2 I don't know what to input for x, y and z to get -$$\nabla$$V since its for all values of r not the ones given.

Don't plug any numbers in for part 2. Leave your answer in terms of $$x, \hat x, y, \hat y, z,$$ and $$\hat z$$.

Also how do you get your letters to adjust to the same size as the symbols, I am new to these forums so I am not quite sure how to do these sort of things yet.

You can copy and paste characters from my signature. To make them italics, highlight them and hit the italics button on the upper left (or hit <Ctrl> I). To make them superscript or subscript, use the X² or X₂ buttons.

There are certain symbols and notation that can't be done using normal characters. But these can be done using LaTeX. There is a link in my signature telling you how to use LaTeX for Physics Forums (PF). The end result is that PF displays the LaTeX as a graphics file. So using LaTeX places a larger load on the PF server (something to keep in mind). Also if you edit an equation and want to preview the results using LaTeX, you might have to hit the Refresh button on your browser after you hit the "Preview Post" button on the PF edit window.

 

[craig.16]

Just had another go at this today in university and worked out where I went wrong. Thanks for the reply anyway.