If the electrostatic potential were Igiven by $$\phi = {\phi _0}$$ (x2 + y2 + z2), where $${\phi _0}$$ is constant, then the charge density giving rise to the above potential would be
A. zero
B. $$ - 6\,{\phi _0}{\varepsilon _0}$$
C. $$ - 2\,{\phi _0}{\varepsilon _0}$$
D. $$ - \frac{{6\,{\phi _0}}}{{{\varepsilon _0}}}$$
Answer: Option B
A. $$\frac{2}{{{\mu _0}}}\left( {x{\bf{\hat i}} + y{\bf{\hat j}}} \right)$$
B. $$ - \frac{2}{{{\mu _0}}}\left( {{\bf{\hat i}} + {\bf{\hat J}}} \right)$$
C. $$ - \frac{2}{{{\mu _0}}}\left( {{\bf{\hat i}} - {\bf{\hat j}}} \right)$$
D. $$\frac{2}{{{\mu _0}}}\left( {x{\bf{\hat i}} - y{\bf{\hat j}}} \right)$$
A. 0.033 μm
B. 0.330 μm
C. 3.300 μm
D. 33.000 μm
A. $${\bf{\hat z}}k$$
B. $${\bf{\hat x}}k\sin \alpha + {\bf{\hat y}}k\cos \alpha $$
C. $${\bf{\hat x}}k\cos \alpha + {\bf{\hat y}}k\cos \alpha $$
D. $$ - {\bf{\hat z}}k$$
A. vp = vg
B. vp = $${\text{v}}_{\text{g}}^{\frac{1}{2}}$$
C. vp vg = c2
D. vg = $${\text{v}}_{\text{p}}^{\frac{1}{2}}$$
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