For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$ where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$ that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is
A. $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{r}$$
B. $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{r}$$
C. $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{{{r^2}}}$$
D. $$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{{{r^2}}}$$
Answer: Option D
This Question Belongs to Engineering Physics >> Electromagnetic Theory
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 = $${\text{v}}_{\text{p}}^{\frac{1}{2}}$$
Join The Discussion