2
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For figure 2 in the above diagram, I derived the following equation:

$$\\E_{o}= E_{1} + E_{2}$$ Where $\\E_1$ and $\\E_2$ are the net electric fields or dielectric 1 and 2, respectively (i.e., $\\E_{1/2}= E - E_p$) And since the net electric field is positive in both cases, the above equation should be correct.

Taking $\\E_1 = \sigma/(\epsilon*k_1)$ and $\\E_2=\sigma/(\epsilon*k_2)$, I'm getting the final equation $$\\C_o*k_1 k_2/ (k_1+k_2)$$

But the answer key says it's $$\\C_o*k_1 k_2/ (k_1-k_2)$$ instead. I don't understand how this is happening. Can someone help, please?

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  • $\begingroup$ Well,what's $E_p$ and $E_{1/2}$...? $\endgroup$ – Nehal Samee Feb 24 '18 at 16:40
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    $\begingroup$ I think your answer is the correct one. Particularly, if you consider the case where $k_1=k_2$, the formula of the answer key blows up; which should not happen. $\endgroup$ – Sahand Tabatabaei Feb 24 '18 at 16:42
  • $\begingroup$ @NehalSamee $\\E_p$ is the internal E of the dielectric due to polarization and $\\E_1/_2$ is $\\E_1$ or $\\E_2$, respectively $\endgroup$ – Antara Kulkarni Feb 24 '18 at 16:57
  • $\begingroup$ How do you transit from “E” to “$C_0$”...? $\endgroup$ – Nehal Samee Feb 24 '18 at 17:00
  • $\begingroup$ I hope you are correct $V=E_1 d + E_2 d$. $$V=\frac {E_0}{k_1} d+ \frac {E_0}{k_2}d$$ ... This leads to your answer ... $\endgroup$ – Nehal Samee Feb 24 '18 at 17:02

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