IV) CONNECTING the FIELD E to the POTENTIAL V
So far, we’ve dealt only with point charges. But if the charges are distributed over a surface or shape, then we must integrate to find V or V.
Possibility 2:
The charge distribution is spherical or cylindrical.
For these cases, the approach to finding V is as follows: E V V
(1) Use Gauss' Law to find E (unless you’re on or outside the surface of a sphere…)
Since V = - E dr (which is the E&M equivalent of U = - F dx), then the next step is to find the potential difference (V) by doing the following:
Find the definite integral between the initial position a and the final position b:
(2) V = -
∫
E dr = Vb - VaSometimes this is the end of it. But other times, you are asked to determine the potential (V) at a point that is a distance r from the center of the sphere/cylinder. This point can be either the outside or inside of the sphere/cylinder.
Here’s what to do for either case:
To find the potential Vr at some distance r that is on or outside the surface of the sphere or cylinder, find the potential difference from r to ∞ (these are the limits of integration).
V = - E dr = V∞ - Vr = 0 - Vr
To find the potential Vr at some distance r that is inside a sphere or cylinder, find the potential difference between R (the radius of the shape, i.e., starting at the surface) and r.
V = -
∫
E dr = Vr - Vsurface ba
R r r
∞
4.1 - Given a charged spherical conductor of radius R and charge +Q, find (a) the potential V at the surface of the sphere, where r = R
(b) the potential at some distance r such that r > R.
(Note - as with (a), this is the same as the potential difference in moving from position r to infinity)
(c) the potential anywhere inside the sphere (r < R)
(d) Plot graphs of E vs r and V vs r
E V
r r
R R
If you are on or outside a sphere, the potential V (as well as the field E) has the same
expression as for a point charge.
The potential inside a conductor is the same as the potential at the surface:
Vinside = Vsurface
R
+Q
Remember, V is the same everywhere along a given line and they’re ALWAYS to the E field lines.
Our answer shows something that is true for any conductor (regardless of its shape): the surface is an equipotential. Otherwise, charge would be moving around freely.
4.2 - Given a solid non-conducting sphere of radius R and having
charge +Q spread uniformly throughout its volume, find both E and V when (a) r > R
(b) r = R (i.e., at the surface of the sphere)
(c) r < R
(d) plot E vs r and V vs r E
r
V
r
R R
Possibility 3: the charge distribution is a ring, arc, disc, or a line segment of charge.
For these cases Gauss' Law cannot be used!! Not to worry! Do the following...
(1) V =
∫
dV = k∫
dq/rThe idea here is that each charge dq on the shape/surface gives rise to a potential dV = k dq/r.
To find the total potential, we integrate (i.e., add up) all the individual contributions from each dq.
Possibility 3 is virtually never used to find V.
Then, if the problem calls for it, the field E can then be found by…
(2) E = -dV/dr* (this is called the potential gradient; same as F = - dU/dx)
4.3 - Find the electric potential for the point P located a distance x away from the center of a ring of charge of radius R. Assume that this point lies along an axis through the center of the ring. Determine the expressions for V & E given
(a) (b)
R • P
r
* The variable for position represents the axis along which the electric field E exists.
So, for example, if a line segment is positioned vertically, then E = -dV/dy.
Or for a ring of charge whose axis extends in the z-direction, E = -dV/dz
a charge Q on the ring a linear charge distribution C/m x
Oh baby! V for a ring (or ring segment) is the same as for a point charge… kQ/r!
(Just be careful about expressing Q & r.)
4.3 (continued)-
(c) Find the expression for the potential at the center of the ring.
(d) Use the answer to (a) to find the expression for the electric field, Ex, at points along the x-axis.
4.4 - Find the net electric potential at the center of a Circular Arc of radius R and charge distribution +that subtends an angle 2.
P R 2
Substitutions:
= dq/ds
dq = dl (2) = s/R
ds = Rd
4.5 - The line segment shown below has a charge distribution +
+ + + + + +
(a) Find the potential V at point P, located a distance x away from the right end of the segment, along the x – axis.
(b) Find the expression for the electric field, Ex, along the x-axis at point P.
4.6 – A thin flat disk, of radius R, carries a uniformly distributed charge Q. Determine the potential V at a point P on the axis of the disk, a distance x from its center.
L
• P x
4.7 – Find the potential at a point P that lies at a distance z on the central axis of a charged disc of radius R and uniform charge density .
But that’s only the first substitution! To be awarded the massive total of 3 points that this would be worth on the AP Exam, we need one more substitution in order to get everything in terms of r!
Now, we need to make the disc into a bunch of extremely thin rings, and then add up (integrate) the contribution from each ring.
Since A = r2, then dA = 2R’dR’, where R’ is the radius of the inner portion of the ring and dR’ is the width of the ring. Here is the next substitution:
V = k/r (2R’dR’) Since r = (z2 + R’2)1/2 we get:
V = k (z2 + R’2)-1/2 (2R’dR’) Result:
V = (2R)[( z2 + R2)1/2 – z]
P
R We need to do integration by substitution twice
in order to solve this problem V = dV = k/r (dq)
= Q/A = dq/dA since it’s a uniform charge density.
This means that dq = dA. So…
V = k/r (dA)
z
R’
r
R 0
Making Sense of Electric Potential, etc.
Point Charges:
Finding Uel, V, V, W ( = qV), K (= qV), and v (using qV0 + K0 = qVf + Kf) is pretty straightforward.
Remember to keep the sign of the charge!
Uniform Field E (including plates & sheets) :
V = Ed remember, Esheet = Eplate =
Rings, Arcs, Discs or Line Segments of Charge:
V = ∫ kdq/r (when charge is given in terms of , use dl instead of dq)
E = -dV/dr
Spheres and Cylinders of Charge:
Can find Va at a given point a that is some separation r from the center by using
V = - ∫ E dr = Vb - Va
where Vb is known:
(i) At a point outside the shape, Vb = V∞ = 0 (ii) At a point inside the shape, Vb = Vsurface
Try to reason out whether V is “+” or “-“ based on whether or not you’re moving along the electric field lines from point a to point b.
Special for Spheres only:
(i) on or outside the surface, E and V are the same as for a point charge!
(ii) inside a conductor, E = 0 & V = Vsurface
(iii) inside an insulator, use Gauss’ Law to get E (you’ll need to use a proportion to find Qenc), and then use V (as above) to find Vinside).
a b
