Week 3: Potential Energy and Potential 111
Now we can differentiate:
Ez = −d
dzkeq
(x^2 + (a−z)^2 )^1 /^2+
d
dzkeq
(x^2 + (a+z)^2 )^1 /^2= − keq(a−z)
(x^2 + (a−z)^2 )^3 /^2− keq(a+z)
(x^2 + (a+z)^2 )^1 /^2(158)
NOW we can letz→0 to find out what the field is on thex-axis (adding and cancelling terms as
necessary, and substitutingpz= 2qain for the dipole moment):
Ez = −^2 keqa
(x^2 +a^2 )^3 /^2= −kepz
(x^2 +a^2 )^3 /^2(159)
Compare this to equation (23)! Hmmm, looks the same^41! And it wasn’tthatdifficult, although it
was certainly more difficult than we might have expected. To see how reallyeasyit was, consider.
We actually just obtained theexactEzfield for all points in space, since the answer is azimuthally
symmetric and we could rotate the answer to tell us the field in planesother than thexzplane! And
theExfield is equally easy to find.
It will turn out that Cartesian coordinates suck in so many ways when doing physics problems.
Physics is if anything naturally spherical or cylindrical – nature is onlyrarely rectilinear. Let’s redo
the potential problem above, but not let’s find the potential at anarbitrary point in spaceinspherical
polar coordinates. Remember, the math section has a lovely little review of Cartesian, Cylindrical
and Spherical coordinate systems – the big three one needs to work with in this course – in case you
have never seen spherical coordinates before (or don’t remember them, effectively the same thing).
(^41) Allowing, of course, for the change in the name of the vertical axis...