InterviewSolution
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InterviewSolution
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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` ? |
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Answer» 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. |
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