The potential of a certain electrostaitc field has the form `varphi = a (x^(2) + y^(2)) + bz^(2)`, where `a` and `b` are constants. Find the magnitude and direction fo the electric field strength vector. What shape have the equipotentail surfaces in the fololowing cases:
(a) `a gt 0, b gt 0`, (b) `a gt 0, b lt 0` ?

Explanation

Given, `varphi = a (x^(2) + y^(2)) + bz^(2)`
So, `vec(E) = -vec(grad) varphi = -[2 ax vec(i) + 2ay vec(j) + 2 bz vec(k)]`
Hence `|vec(E)| = 2 sqrt(a^(2) (x^(2) + y^(2)) + b^(2) z^(2))`
Shape of the equipotential surface :
Put `vec(rho) = x vec(i) + y vec(j)` or `rho^(2) = x^(2) + y^(2)`
Then the quipotential surface has the equaction
`a rho^(2) + b z^(2)` = constant `= varphi`
If `a gt 0, b gt 0` then `varphi gt 0` and teh equaction of the equipotential surface is
`(rho)/(varphi//a) + (z^(2))/(varphi//b) = 1`
which is an ellipose in `rho, z` coordinates. In three dimensions the surface is an elipsoid of revolution with semi-axis `sqrt(varphi//a), sqrt(varphi//a), sqrt(varphi//b)`.
If `agt0, blt0` then `varphi` can be `ge 0`. If `varphi gt 0` then the equaction is
`(rho^(2))/(varphi//a) - (z^(2))/(varphi//|b|) = 1`
This si a single cavity hyperboloid of revolution about `z` axis. If `varphi = 0` then
`a rho^(2) = |b| z^(2) = 0`
or `z = +- sqrt((a)/(|b|)) rho`
is the equaction of a right circular cone.
If `varphi lt 0` then the equaction can be written as
`|b| x^(2) - a rho^(2) = |varphi|`
or `(x^(2))/(|varphi|//|b|) = (rho^(2))/(|varphi|//a) = 1`
This is a two cavity hyperbotold of revolution about z-axis.