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In the arrangement shown in figure dielectric constant `K_1=2` and `K_2=3`. If the capacitance across `P` and `Q` are `C_1` and `C_2` respectively, then `C_1//C_2` wil be (the gaps shown are negligible) `A. `1:1`B. `2:3`C. `9:5`D. `25:24` |
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Answer» Correct Answer - D `C_1=C_(RHS)+ C_(LHS)` `=(K_2epsilon_0(A/2))/d=(K_1epsilon_0(A/2))/d` `(epsilon_0A)/(2d)(K_1+K_2)=(5epsilon_0A)/(2d)` `C_2=(epsilon_0A)/(d-d/2-d/2+(d/2)/K_1+(d/2)/K_2` `=(2epsilon_0A)/d((K_1K_2)/(K_1+K_2))` `=(12epsilon_0A)/(5d)` `:. C_1/C_2=25/24` |
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