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Two identical parallel plate capacitors are connected in one case in parallel and in the other in series. In each case the plates of one capacitors are brought closer by a distance a and the plates of the outer are moved apart by the same distance a. ThenA. total capacitance of first system increasesB. total capacitance of first system decreasesC. total capacitance of second system decreasesD. total capacitance of second system remains costant. |
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Answer» Correct Answer - A::D a.,d. When capacitors are connected in parallel, initial capacitance is `C=2epsilon_(0)A//d`. After distance between the plates is changed, the capacitance becomes `C=(epsilon_(0)A)/(d+a)+(epsilon_(0)A)/(d-a)` which is greater than initial one. hence option (a) is correct and option (b) is incorrect. When capacitors are connected in series, then `(1)/(C)=(2d)/(epsilon_(0)A)` After the distance between the plates is changed, `(1)/(C)=(d+a)/(epsilon_(0)A)+(d-a)/(epsilon_(0)A)=(2d)/(epsilon_(0)A)` |
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