The capacitance of a parallel plate capacitor with plate area A and separation d, is C in vacuum .The space between the plates is filled with two wedges of dielectric constant \(K_1 \) and \(K_2\) respectively . Find the capacitance of the resulting capacitor in \(\mu F\)

\(Where\quad \\ A=4\pi \quad ㎡\\ d=\frac { { 10 }^{ -3 } }{27 } \quad m\\ { K }_{ 1 }\quad =\quad 5\\ { K }_{ 2 }\quad =\quad 2\\ 4\pi { \epsilon }_{ 0 }=\frac { 1 }{ 9.{ (10 }^{ 9 }) } \)

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