8—Multivariable Calculus 218is
E~(x,y,z) =kq
r^2ˆrwhereˆris the unit vector pointing away from the origin andris the distance to the origin. This vector can be
written as a gradient. Because thisE~is everywhere pointing away from the origin, it’s everywhere perpendicular
to the sphere centered at the origin.
E~=−gradkq
r
You can verify this a several ways. The first is to go straight to the definition of a gradient. (There’s a blizzard
of minus signs in this approach, so have a little patience. It will get better.) This function is increasing most
rapidly in the direction moving toward the origin. ( 1 /r) The derivative with respect to distance in this direction
is−d/dr, so−d/dr(1/r) = +1/r^2. The direction of greatest increase is along−ˆr, so grad(1/r) =−ˆr(1/r^2 ).
But the relation to the electric field has another− 1 in it, so
−gradkq
r= +rˆkq
r^2There’s got to be a better way.
Yes, instead of insisting that you move in the direction in which the function isincreasing most rapidly,
simply move in the direction in which it is changing most rapidly. The derivative with respect to distance in that
direction is the component in that direction and the plus or minus signs take care of themselves. The derivative
with respect torof ( 1 /r) is− 1 /r^2. That is the component in the directionˆr, the direction in which you took
the derivative. This says grad(1/r) =−ˆr(1/r^2 ). You get the same result as before but without so much fussing.
This also makes it look more like the familiar ordinary derivative in one dimension.
Still another way is from the “Stallone-Schwarzenegger” brute force school of computing. Express everything
in rectangular coordinates and do the partial derivatives using Eqs. ( 9 ) and ( 8 ).
(
∂(1/r)
∂x
)
y,z=
(
∂(1/r)
∂r)
θ,φ(
∂r
∂x)
y,z=−
1
r^2∂
∂x√
x^2 +y^2 +z^2 =−1
r^2x
√
x^2 +y^2 +z^2Repeat this foryandzwith similar results and assemble the output.
−gradkq
r=
kq
r^2xˆx+yˆy+zzˆ
√
x^2 +y^2 +z^2=
kq
r^2~r
r=
kq
r^2ˆr