A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
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Step 1: Initial capacitance
Let the initial distance between the plates be d.
Let A be the area of each plate.
Capacitance of the parallel plate capacitor is given by:
Step 2: Capacitance after inserting dielectric
Let k be the dielectric constant.
As the distance between plates is reduced to half, new distance is d′=2d
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