phys quiz...

[Sarahisme]

1)

True or false:

(a) If the electric field is zero in some region of space, the electric potential must also be zero in that

region.

(b) If the electric potential is zero in some region of space, the electric field must also be zero in that

region.

© If the electric potential is zero at a point, the electric field must also be zero at that point.

(d) Electric field lines always point toward regions of lower potential.

(e) The value of the electric potential can be chosen to be zero at any convenient point.

(f) In electrostatics, the surface of a conductor is an equipotential surface.

 

2)

If the voltage across a parallel-plate capacitor is doubled, will its capacitance (a) double, (b) drop by half

or © remain the same?

 

 

 

My answers would be :

1)

a) TRUE

b) TRUE

c) TRUE

d) FALSE

e) FALSE

f) TRUE

 

2) (b)

 

how'd i go?

 

Sarah

[Johnny5]

1)

True or false:

(a) If the electric field is zero in some region of space' date=' the electric potential must also be zero in that

region.

 

[/quote']

 

Without thinking at all, I think the answer is false.

 

Now, I'll just use math, to answer it for sure.

 

Here is the definition of electric field:

 

E = 1/4pe₀ integral rdt R^/|R|²

 

Ok now, the following equivalence can be shown by purely mathematical means:

 

R^/|R|² = -Ñ(1/|R|)

 

So that we can rewrite the electric field E as:

 

E = 1/4pe₀ integral rdt -Ñ(1/|R|)

 

Now, the integral is with respect to the source coordinate frame S`, and the gradient operator involves partial derivatives on the field points (x,y,z). So if we imagine the integral to have been completed over one moment in time, we can pull out the del operator like so:

 

E = Ñ {1/4pe₀ integral rdt -(1/|R|)}

 

Where the del operator now acts upon everything inside the curly braces.

 

Now, that which is interior to the braces is a scalar function rather than a vector function. It is customary to pull out the - sign, and label the interior scalar function v, for voltage, like so:

 

E = -Ñ v

 

Where v = 1/4pe₀ integral rdt (1/|R|)

 

v is also called "electric potential."

 

The most important thing about v, is that it is a scalar function rather than a vector function, which makes it a bit easier to work with.

 

Now, suppose that you have chosen points in the reference frames S,S` to have rectangular coordinates (x,y,z), (x`, y`, z`) respectively.

 

Now, if your blob of matter has rest frame S`, and is at rest in frame S, and there are no other electrically charges bits of matter anywhere, your electric field value (at any randomly chosen field point (x,y,z) in the frame) will remain constant for short durations of time, and will be given by - gradient of the voltage at the field point (x,y,z). So here is what the gradient operator is, in rectangular coordinates:

 

 

¶ /¶x i + ¶ /¶y j + ¶ /¶z k

 

Where i,j,k represent unit vectors.

 

As you can see, the partial derivatives, are derivatives (changes) in the location of the field point, not changes in a source point. This is conceptually very important, when it comes to electrodynamics. Right now, we have a blob of electrically charged matter at rest in reference frame S, and the superfluous electrons on it are also at rest in frame S. And since S` is the rest frame of the blob, there is no relative motion between frames S, S`.

 

So the Electric field E, at the field point (x,y,z) due solely to the existence of the blob, is constant in time.

 

So, to answer your question, suppose that the electric field E, at a field point happens to be zero. What does that say about the voltage v?

 

Well in such a case we have:

 

E = 0 = -Ñ v

 

Now, here is your question one:

 

1)

True or false:

(a) If the electric field is zero in some region of space, the electric potential must also be zero in that

region.

 

If we can find a counterexample to the statement, then the answer is false.

 

From calculus, you know that the derivative of any constant is zero.

 

So suppose that the electric potential at the field point is 4. The partial derivative of 4 with respect to x is zero, the partial derivative of 4 with respect to y is zero, and the partial derivative of 4 with respect to z is zero. Thus, if v=4 then we have the following true statement:

 

-Ñ 4 = ¶4 /¶x i + ¶4 /¶y j + ¶4 /¶z k = 0

 

So, the voltage at the field point can be a nonzero constant, and the electric field there will be zero. So it isn't necessary that the potential at the field point be zero, in order for the electric field there to be zero. What is necessary for the E field to be zero there, is for the potential to be a spatio-temporal constant.

 

I have just read Dr. Swanson's post, and that is what allows you to intuitively know that the answer to question one is false, without doing all the mathematics. Also, you can use that fact to reduce the number of mesh equations you have, when you are working with Kirchoff's laws, in order to generate equations in v,i. You can set the ground to be zero, wherever you want in a circuit, and things will still work out.

 

I doubt you want to get that involved here. Here, you simply want to know a quick way to get at the right answer. But if you already know electrodynamics, then you will have run through the theory at least once, and should already know that you can add an arbitrary constant to the voltage, and your electric field at a point doesn't change.

 

The reason is quite mathematical.

 

Suppose that the following scalar function is the potential at a point:

 

V(x,y,z) = e/4pe₀r

 

Where r = (x^2+y^+z^2)^1/2

 

The electric potential above, is the potential due to a single electron, with electric charge e, located at the origin of reference frame S.

 

You can compute the electric field at an arbitrary field point, by taking the gradient. The answer you get will be the same as if the electric potential was:

 

 

V(x,y,z) = e/4pe₀r + C

 

Where C is any constant.

 

Which is what Dr. Swanson alluded to.

 

So after all that, you can be certain that the answer to question a... is false.

 

 

I see that Dr. Mattson has answered the others, so if I have nothing further to add, I will leave things here.

 

Without reading his answer to b, you should be able to figure out the answer.

 

Suppose that v=0. Take your partials, and you will get E=0.

 

The answer to c is not so easy to understand as the answer to b. Right now I am trying to think of a realistic potential function v(x,y,z), which has a value zero at some point, yet the electric field at the point is nonzero.

 

I have already read Dr. Mattsons answer, and he got me thinking of a paraboloid.

 

Here is the formula for it:

 

z = x²/a² + y²/b²

 

The above formula, leads to the shape of a satellite dish, a circular paraboloid.

 

As you can see from the scalar function, when x=y=0, z is also equal to zero.

 

So suppose you have this, due to some charge distribution somewhere:

 

 

v = x²/a² + y²/b²

 

You can see that the potential function is zero, when x=y=0.

 

Let's look at the gradient of v.

 

Ñv = 2x/a² + 2y/b²

 

As it turns out, this implies that when x=y=0, that the electric field is zero too.

 

But you should now be able to see the following:

 

Had the exponents of x,y been 1, instead of zero, then after the partial derivatives were taken, we would have had a constant for an answer. So, suppose that you have the following potential function:

 

v(x,y,z) = 3x+2y+5z

 

Clearly, you have v(0,0,0) = 3*0+2*0+5*0=0

 

So at the point (0,0,0) in a frame, the voltage is zero, but now take the partials to obtain:

 

gradient v = 3i^+2j^+5k^

 

Which is clearly nonzero everywhere.

 

I'm not saying that the above function for v corresponds to any realistic configuration of charge distribution throughout the universe, but it does make Dr. Mattson's point, which is that there are mathematical functions which constitute counteraxamples to part © here.

 

 

 

 

¶ alt-0182

 

Ñ alt-0209

[swansont]

The electric field is the negative gradient of the potential. Which means you can add a constant to the potential and still get the same field - the choice of potential is arbitrary at one point.

 

So you might want to revisit your answers to a-e.

 

This might help.

[Tom Mattson]

how'd i go?

 

Sarah

 

Not so hot.

 

Let's take a look...

 

1)

True or false:

(a) If the electric field is zero in some region of space' date=' the electric potential must also be zero in that

region.

[/quote']

 

For simplicity consider an electric field in the x-direction only. Then E=-dV/dx. If E=0, it does not mean that V=0. It means that the slope of V(x) is 0.

 

(b) If the electric potential is zero in some region of space, the electric field must also be zero in that

region.

 

As long as that region is more than a single point, then this is true. In fact this would be true if V were any constant, because d(constant)/dx=0. So you got this one right.

 

© If the electric potential is zero at a point, the electric field must also be zero at that point.

 

Think about this. If the potential V(x) is zero at some point then that just means it has an x-intercept at that point. Is it necessarily the case that the function is neither increasing nor decreasing there? No, of course not.

 

(d) Electric field lines always point toward regions of lower potential.

 

Look at any diagram of electric field lines. The lines always go from positive charges towards negative charges. What does that tell you?

 

(e) The value of the electric potential can be chosen to be zero at any convenient point.

 

No matter what V(x) is, you can always modify it by adding a constant c. This is because V(x) and V(x)+c give the same electric field. Since it is the field that is measureable, that is all we care about. So the two potentials above are physically equivalent. And c can always be chosen to be -V(x₀) for an arbitrary x=x₀. What does that tell you?

 

(f) In electrostatics, the surface of a conductor is an equipotential surface.

 

You got this one right.

 

2)

If the voltage across a parallel-plate capacitor is doubled, will its capacitance (a) double, (b) drop by half

or © remain the same?

 

Capacitance depends on geometry only. Not on q or V.

[Sarahisme]

lol, so which ones did i actually get right? just question 2?

[Sarahisme]

so let me try again, these are my revised answers....

 

1

a) F

b) T

c) F

d) T

e) T

f) T

 

2) c

 

please tell me thats right? :S

[Tom Mattson]

lol, so which ones did i actually get right? just question 2?

 

On Take One you got 1b and 1f right. You're looking good now.

 

But more importantly, do you understand the reasoning process?

[Johnny5]

On Take One you got 1b and 1f right. You're looking good now.

 

But more importantly' date=' do you understand the reasoning process?[/quote']

 

Tom, what would you say to my answer of part c, it's a bit long winded. Here it is:

 

The answer to c is not so easy to understand as the answer to b. Right now I am trying to think of a realistic potential function v(x' date='y,z), which has a value zero at some point, yet the electric field at the point is nonzero.

 

I have already read Tom's answer, and he got me thinking of a paraboloid.

 

Here is the formula for it:

 

z = x2/a2 + y2/b2

 

The above formula, leads to the shape of a satellite dish, a circular paraboloid.

 

As you can see from the scalar function, when x=y=0, z is also equal to zero.

 

So suppose you have this, due to some charge distribution somewhere:

 

 

v = x2/a2 + y2/b2

 

You can see that the potential function is zero, when x=y=0.

 

Let's look at the gradient of v.

 

Ñv = 2x/a2 + 2y/b2

 

As it turns out, this implies that when x=y=0, that the electric field is zero too.

 

But you should now be able to see the following:

 

Had the exponents of x,y been 1, instead of zero, then after the partial derivatives were taken, we would have had a constant for an answer. So, suppose that you have the following potential function:

 

v(x,y,z) = 3x+2y+5z

 

Clearly, you have v(0,0,0) = 3*0+2*0+5*0=0

 

So at the point (0,0,0) in a frame, the voltage is zero, but now take the partials to obtain:

 

gradient v = 3i^+2j^+5k^

 

Which is clearly nonzero everywhere.

 

I'm not saying that the above function for v corresponds to any realistic configuration of charge distribution throughout the universe, but it does make Tom's point, which is that there are mathematical functions which constitute counteraxamples to part © here.[/quote']

 

Specifically, the equation for a plane works.

 

But is there any configuration of charge which gives a plane for the potential function?

 

I couldn't think of any.

[Tom Mattson]

Tom' date=' what would you say to my answer of part c, it's a bit long winded. Here it is:

[/quote']

 

Sorry, I don't have the patience for all that. You can use my answer to part (e) to immediately write down a potential function that is zero at a point where the electric field is nonzero.

 

Start with the Coulomb potential V®=-ke/r and add a term ke/r₀. So V®=-ke/r+ke/r₀. This gives the exact same E-field as the original V®, and it is zero at r=r₀ (where E is nonzero).

[Johnny5]

You can use my answer to part (e) to immediately write down a potential function that is zero at a point where the electric field is nonzero.

 

Start with the Coulomb potential V®=-ke/r and add a term ke/r₀. So V®=-ke/r+ke/r₀. This gives the exact same E-field as the original V®' date=' and it is zero at r=r[sub']0[/sub] (where E is nonzero).

 

Well lets see, here is your answer to part e:

 

(e) The value of the electric potential can be chosen to be zero at any convenient point.

 

No matter what V(x) is' date=' you can always modify it by adding a constant c. This is because V(x) and V(x)+c give the same electric field. Since it is the field that is measureable, that is all we care about. So the two potentials above are physically equivalent. And c can always be chosen to be -V(x0) for an arbitrary x=x0. What does that tell you?[/quote']

 

Actually this part here makes perfect sense:

 

Start with the Coulomb potential V®=-ke/r and add a term ke/r0. So V®=-ke/r+ke/r0. This gives the exact same E-field as the original V®, and it is zero at r=r0 (where E is nonzero).

 

Let V® = -ke/r.

 

Now compute the gradient of V®.

 

Ñ V® = ¶V /¶r r

= ¶(-ke/r) /¶r r

 

= -ke¶(1/r) /¶r r

 

= -ke(-r^-2) r

 

= ke/r²r

 

Since

E = -Ñ v

 

It follows that:

 

E = - ke/r²r

 

Which is the electric field of an electron, at the origin of a reference frame.

 

The force on a positively charged test charge, with charge Q, is given by QE, hence we have:

 

F = - kQe/r²r

 

The negative sign implies the force is attractive, since Q, and e are positive.

 

Ok so we have the potential of an electron, as a function of r, namely:

 

V® = -ke/r

 

Now, suppose we add the constant ke/r₀ to V®, to obtain:

 

 

V® = -ke/r + ke/r₀

 

When we take the partial with respect to r, we get the same gradient function as before, hence the same electric field.

 

But clearly, when r=r0 the potential is zero, that is:

 

If V® = -ke/r + ke/r₀

then V(r₀)=0

 

So r0 is a point in the field where the potential is zero, as you can see.

 

It cannot be the origin, because that is division by zero error, unless the charge unit e, is zero inside the electron.

[Sarahisme]

On Take One you got 1b and 1f right. You're looking good now.

 

But more importantly' date=' do you understand the reasoning process?[/quote']

 

yeah i think i do understanding, well so long as i got all the answers right i suppose... because i think i have now, havent i? :S