All Problems

Constant Electric Field in a Vacuum

Problem 3.37

The potential of a certain electrostatic field has the form φ=a(x2+y2)+bz2,\varphi=a\left(x^{2}+y^{2}\right)+b z^{2}, where aa and bb are constants. Find the magnitude and direction of the electric field strength vector. What shape have the equipotential surfaces in the following cases: (a) a>0,b>0a>0, b>0; (b) a>0,b<0?a>0, b<0 ?

Reveal Answer
 3.37. E=−2(axi+ayj+bzk),E=2a2(x2+y2)+b2z2 \text { 3.37. } \mathbf{E}=-2(a x \mathbf{i}+a y \mathbf{j}+b z \mathbf{k}), \quad E=2 \sqrt{a^{2}\left(x^{2}+y^{2}\right)+b^{2} z^{2}} (a) An ellipsoid of revolution with semiaxes φ/a\sqrt{\varphi / a} and φ/b\sqrt{\varphi / b}. In the case of φ>0,\varphi>0, a single-cavity hyperboloid of revolution; when φ=0,\varphi=0, a right round cone; when φ<0,\varphi<0, a two-cavity hyperboloid of revolution.
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