InterviewSolution
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InterviewSolution
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Between the plates of a parallel plate condenser, a plate of thickness `t_(1)` and dielectric constant `k_(1)` is placed. In the rest of the space, there is another plate of thickness `t_(2)` and dielectric constant `k_(2)`. The potential difference across the condenser will beA. `(Q)/(A epsilon_(0))((t_(1))/(k_(1))+(t_(2))/(k_(2)))`B. `(epsilon_(0)Q)/(A)((t_(1))/(k_(1))+(t_(2))/(k_(2)))`C. `(Q)/(A epsilon_(0))((k_(1))/(t_(1))+(k_(2))/(t_(2)))`D. `(epsilon_(0)Q)/(A) (k_(1)t_(1) + k_(2)t_(2))` |
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Answer» Correct Answer - A Potential difference across the condenser `V = V_(1) + V_(2) = E_(1)t_(1) + E_(2)t_(2) = (sigma)/(K_(1)epsilon_(0)) + (sigma)/(K_(2)epsilon_(0)) t_(2)` `rArr V = (sigma)/(epsilon_(0)) ((t_(1))/(K_(1))+(t_(2))/(K_(2))) = (Q)/(A epsilon_(0)) ((t_(1))/(K_(1))+(t_(2))/(K_(2)))` |
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