Find the equivalent capacitance between and , , where is area of plates with
Assume each conducting plate is having same dimensions and neglect the thickness of the plate.
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Mark the plates as following:
This arrangement of plates can be converted into a circuit of capacitors as shown below:
Notice how the plates are connected according to the first diagram!
This is just a simple capacitor circuit, which considers C 3 4 = C 3 2 = C 2 1 = C 3 1 = 7 μ F = C and so, the capacitance, C 4 2 = 2 C .
Solving this network, we get:
C n e t = 7 1 1 C = 1 1 μ F