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Electric field question(half worked already)

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Homework Statement



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Homework Equations





The Attempt at a Solution



the electric field that Q is subjected to : E = q/(4piεor^2) ...... i think

force that Q is subjected to :

i substitute k here for 1/(4piεo) to get F = (kQq)/(r^2).... i know the capital Q is supposed to be there, but dont know how it is fitted in algebraically.

then its just dotted into some angles for the vector math

r = √ (x^2 + a^2)

Force exerted on Q, F = (kQ|q|)/(x^2 + a^2) ( cosϑ i - sinϑ j ) + (kQ|-q|)/(x^2 + a^2) ( - cosϑ i - sinϑ j )

= -2(kQ|q|)/(x^2 + a^2) sin j = -2(kQ|q|)/(x^2 + a^2) * (a/√ (x^2 + a^2) )


work done in moving Q from (0,0) to (x,0)

not sure how to tackle this one.. I have an idea though W = ∫ F . dr or F . Δ r

can i just multiply my above result for the force into the change in x from the origin to where the point sits at (x,0)

something like (F . 0) + (F . x) ?


could someone give me a hand with what my start, i know the math looks a little shaky, but i think what i have up is very close to correct for the first two parts of the question.
 
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Answers and Replies

  • #2
collinsmark
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The Attempt at a Solution



the electric field that Q is subjected to : E = q/(4piεor^2) ...... i think

force that Q is subjected to :

i substitute k here for 1/(4piεo) to get F = (kQq)/(r^2).... i know the capital Q is supposed to be there, but dont know how it is fitted in algebraically.

then its just dotted into some angles for the vector math

r = √ (x^2 + a^2)

Force exerted on Q, F = (kQ|q|)/(x^2 + a^2) ( cosϑ i - sinϑ j ) + (kQ|-q|)/(x^2 + a^2) ( - cosϑ i - sinϑ j )

= -2(kQ|q|)/(x^2 + a^2) sin j = -2(kQ|q|)/(x^2 + a^2) * (a/√ (x^2 + a^2) )
Okay, good! :approve: It looks like found the expression for the force's magnitude. But don't forget about direction. That's important.

But let's go back to the electric field. What is the relationship between the electric field at a given point in space, and the force on a charge Q placed at the same point in space? Hint: the answer is in the very definition of electric field. My reason for asking/saying all of this is that it's trivial to find the electric field if you already happen to know the force, and vice versa.
work done in moving Q from (0,0) to (x,0)

not sure how to tackle this one.. I have an idea though W = ∫ F . dr or F . Δ r
can i just multiply my above result for the force into the change in x from the origin to where the point sits at (x,0)

something like (F . 0) + (F . x) ?
No, you can't do it that way. The force varies in between 0 and x, So what you're proposing isn't valid.

However, the W = ∫ F · dr formula is valid, and is one possible approach.
could someone give me a hand with what my start, i know the math looks a little shaky, but i think what i have up is very close to correct for the first two parts of the question.
There's actually more than one way to solve for the work done. The way I would do it is to calculate the difference in electric potential between points 0 and x; and then find the work done from there. But I'm not sure if you've covered electric potential in your coursework yet (if not, you will learn it soon :smile:).

But let's go back to the W = ∫ F · dr approach. This will work too. And for this particular problem, it turns out to be quite easy using this approach. The thing that makes it easy is to recall the properties of the dot product. What is the dot product between two parallel vectors? And perpendicular vectors? What is the direction of dr as you go from 0 to x? What is the direction of the force? :wink:
 
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There's actually more than one way to solve for the work done. The way I would do it is to calculate the difference in electric potential between points 0 and x; and then find the work done from there. But I'm not sure if you've covered electric potential in your coursework yet (if not, you will learn it soon :smile:).
The professor hinted at it, but we are not quite there yet!

But let's go back to the W = ∫ F · dr approach. This will work too. And for this particular problem, it turns out to be quite easy using this approach. The thing that makes it easy is to recall the properties of the dot product. What is the dot product between two parallel vectors? And perpendicular vectors? What is the direction of dr as you go from 0 to x? What is the direction of the force? :wink:
What is the dot product between two parallel vectors
X . Y = |X||Y|cos(0) = |X||Y|(1)

And perpendicular vectors?
X . Y = |X||Y|cos(pi/2) = 0

What is the direction of dr as you go from 0 to x?
positive "i" in "vector speak", right?

What is the direction of the force?
negative "i", from what I calculated above.





so I will take a stab at this, vectors and integrations are one thing I do not have much experience in:

∫ -2(kQ|q|)/(x^2 + a^2) * (a/√ (x^2 + a^2) ) . dr

and I believe everything is constant besides dr, so it may come out of the integral

= -2(kQ|q|)/(x^2 + a^2) * (a/√ (x^2 + a^2) ) ∫ dr

= -2(kQ|q|)/(x^2 + a^2) * (a/√ (x^2 + a^2) ) r + C

and wouldn't r just be the distance that Q sits away from the origin? This looks too much like F . Δ r, I must have made a mistake somewhere because my integral didnt involve any dot products.
 
  • #4
collinsmark
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The professor hinted at it, but we are not quite there yet!
What is the dot product between two parallel vectors
X . Y = |X||Y|cos(0) = |X||Y|(1)
And perpendicular vectors?
X . Y = |X||Y|cos(pi/2) = 0
Yes! :approve: Very nice. The dot product is multiplication of one of the vector's magnitude, multiplied by the magnitude of the projection of the other vector onto the first. The dot product of perpendicular vectors is zero (the projection is always zero).
What is the direction of dr as you go from 0 to x?
positive "i" in "vector speak", right?
Yes, that's right. :approve: the direction is along the positive x-axis.
What is the direction of the force?
negative "i", from what I calculated above.
Oooh. Not so right. :frown: Remember, the top charge has a positive sign (+q) and the bottom charge has a negative sign (-q). Try that one again.
I must have made a mistake somewhere because my integral didnt involve any dot products.
The first mistake is about the direction of the force. Make that correction first. It seems you must have made the correct assumptions in your original work, but then forgot to tack on the final unit vectors.

[tex] \vec F = ( \mbox{sum of force components along the x-direction} ) \hat \imath + ( \mbox{sum of force components along the y-direction} ) \hat \jmath [/tex]

Notice the unit vectors, [tex] \hat \imath [/tex] and [tex] \hat \jmath [/tex] above.

/////////////////////////////

Now a quick review of the dot product, in Cartesian coordinates.

Suppose we have two vectors, [tex] \vec a [/tex] and [tex] \vec b [/tex].

[tex] \vec a = a_x \hat \imath + a_y \hat \jmath +a_z \hat k [/tex]

[tex] \vec b = b_x \hat \imath + b_y \hat \jmath +b_z \hat k .[/tex]

Then the dot product is,

[tex] \vec a \cdot \vec b = a_x b_x + a_y b_y + a_z b_z [/tex]

Notice the pesky unit vectors go away, and the result is a scalar.
 
  • #5
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the force is negative j right? :uhh:

before I did the transformation back to x, the force looked like this

f = -2(kQ|q|)/(x^2 + a^2) sin j

with a - sinj in there, which makes sense, because q repels and - q attracts.

plugging into the integral and using dr i

w = ∫ -2(kQ|q|)/(x^2 + a^2) sin j . dr i

so the vectors should go away here

w = ∫- 2(kQ|q|)/(x^2 + a^2) sin dr

preforming the integral

= 2(kQ|q|)/(x^2 + a^2) cos*r + C

is that on the right track? Thanks so much for helping me
 
  • #6
collinsmark
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the force is negative j right? :uhh:

before I did the transformation back to x, the force looked like this

f = -2(kQ|q|)/(x^2 + a^2) sin j

with a - sinj in there, which makes sense, because q repels and - q attracts.
Not [tex] \sin \hat \jmath [/tex], just [tex] \hat \jmath [/itex]

So you have

[tex] \vec F = 0 \hat \imath \ - 2k \frac{Qq}{x^2 + a^2} \left( \frac{a}{\sqrt{x^2 + a^2}} \right) \hat \jmath \ + 0 \hat k [/tex]
plugging into the integral and using dr i

w = ∫ -2(kQ|q|)/(x^2 + a^2) sin j . dr i

so the vectors should go away here

w = ∫- 2(kQ|q|)/(x^2 + a^2) sin dr
Be careful.

[tex] \vec{dr} = dx \hat \imath \ + 0 \hat \jmath + \ 0 \hat k [/tex]

Now try taking the dot product again. :wink:

[Edit: fixed a formula.]
 
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  • #7
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Not [tex] \sin \hat \jmath [/tex], just [tex] \hat \jmath [/itex]

So you have

[tex] \vec F = 0 \hat \imath \ - 2k \frac{Qq}{x^2 + a^2} \left( \frac{a}{\sqrt{x^2 + a^2}} \right) \hat \jmath \ + 0 \hat k [/tex]

Be careful.

[tex] \vec{dr} = dx \hat \imath \ + 0 \hat \jmath + \ 0 \hat k [/tex]

Now try taking the dot product again. :wink:

[Edit: fixed a formula.]

dot product is now zero, work done is zero right?
 
  • #8
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dot product is now zero, work done is zero right?
Correct! :approve:

[Edit: Now take a step back and consider the direction of the vectors involved (i.e. how parallel/perpendicular they are to each other) and how that relates to the dot product.]
 
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