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That brings us, then, to problem number four--
this is a semi-classic, 24-68.
I have an insulating arc, which is half a circle, and I
put a charge here on the arc at a uniform density.
I have a total charge of Q on the arc, and the radius
of the arc is a--
it's only half an arc.
Lambda, if you want to express it how many coulombs there are
per meter, lambda would be Q divided by half the
circumference, so that it's pi a.
Now you're being asked: what is the potential at that point
P, right at the center of that arc?
As I said earlier, the potential is a scalar--
it's not a vector.
The advantage of scalars is that you don't have to take
into account directions--
they have no direction, so all you have to do is just add up
the individual potentials from each one of those little
sections over the arc, and that's exactly what
I'm going to do.
I'm going to carve out here a little section, dx.
That little section dx contains a charge dq, which is
lambda times dx, which is therefore Q divided
by pi a times dx.
That is the little amount of charge here.
The potential at the point P as a result of this charge, if
I assume that infinity is 0, that potential,
which I will call dVp--
it's only a teeny weeny little bit due to this section dx.
It therefore equals Q divided by pi a divided by 4
pi a epsilon 0.
Remember, it's 4 pi epsilon 0 r--
r is the distance from the charge to that point P, and
we've seen it in our previous problems--
so that's why this a is there times the x.
Now I have to add all these elements of dx, so
that makes it easy--
I do an integral, and the integral has to be
done along the arc.
That will give me that V p is Q divided by pi a divided by
this times the integral along the arc of dx.
The integral along the arc of dx itself is pi a, so
I lose the pi a.
I get Q divided by 4 pi epsilon 0 times a--
that would be the potential at this point
P due to this half-arc.
Now you are being asked what is the electric
field at point P?
That's harder, because the electric field is a vector, so
now if you take this little element, the positive charge
here would have a small e vector is this direction.
But if you take the little element here, that would have
e vector in this direction, and if you take the small
element here, that would have an e vector in this direction.
You will have to add up all these e vectors to get the net
e vector, and I will help you a little bit: the net e
vector, I claim, must be in the direction e net.
That goes right through the center of this arc.
The reason for that is that it's clear that the y
components--
if I called these the y components--
will cancel because of the symmetry of the problem.
What I have here below this horizontal line is exactly
what I have here, so you don't even have to worry about the y
components--
all you have to do is calculate all the x
components, due to all these little elements.
Then you have to do an integral over the whole arc,
forgetting the y components, and out comes the net e
vector, which will be, I guarantee
you, in this direction.