Discussion
At the uncharged interface between two linear dielectrics with permitivities \(\epsilon_{1}\) and \(\epsilon_{2}\), electric fields \(\overline{E_{1}}\) and \(\overline{E_{2}}\) make angles \(\theta_{1}\) and \(\theta_{2}\) with the normal. The ratio \(\epsilon_{1} / \epsilon_{2}\) is equal to :
Answer: A
The ratio is \(\epsilon_{1}/\epsilon_{2} = \tan\theta_{1}/\tan\theta_{2}\).
At an uncharged interface the tangential component of E and the normal component of D are continuous:
\[ \begin{aligned} E_{1}\sin\theta_{1} &= E_{2}\sin\theta_{2} \\ \epsilon_{1}E_{1}\cos\theta_{1} &= \epsilon_{2}E_{2}\cos\theta_{2} \end{aligned} \]
Dividing the first equation by the second gives \(\tan\theta_{1}/\epsilon_{1} = \tan\theta_{2}/\epsilon_{2}\), the law of refraction of field lines.
The sine ratio is Snell's law for light, which involves refractive indices rather than permittivities. The cosine and cotangent ratios do not follow from the boundary conditions; the cotangent ratio is in fact the inverse of the correct answer.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development