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Conductors and Dielectrics in an Electric Field

Problem 3.85

Solve the foregoing problem for the case when half the gap is filled with the dielectric in the way shown in Fig. \(3.13 .\)

Reveal Answer
 3.85. (a) E1=E2=E0,D1=ε0E0,D2=εD1; (b) E1=E2==2E0/(ε+1),D1=2ε0E0/(ε+1),D2=εD1\begin{array}{l} \text { 3.85. (a) } E_{1}=E_{2}=E_{0}, D_{1}=\varepsilon_{0} E_{0}, D_{2}=\varepsilon D_{1} ; \text { (b) } E_{1}=E_{2}= \\ =2 E_{0} /(\varepsilon+1), D_{1}=2 \varepsilon_{0} E_{0} /(\varepsilon+1), D_{2}=\varepsilon D_{1} \end{array}
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