Force of attraction between oppositely charged plates of a capacitor....
Asked by harinderamit1234
25th September 2018,
6:18 PM
Answered by Expert
Answer:
If the dimensions of parallel plates are much grater than the separation distance of plates,
we can assume the electrostatic field experienced by a plate due to charges on other plate is
equivalent to field due to charged infinite plane sheet.
Hence elctrostatic field experienced by a plate due to charges on other plate is σ/(2εo) ,
where σ = Q/A is surface charge density , Q is the charge on plate and A is area of plate.
Hence force of attraction = Q×E = Q2/(2×A×εo) ..................(1)
if C is capacitance of the given capacitor and V is applied voltage, then C = Q/V or Q = C×V ..............(2)
substituting for Q in eqn.(1), force of attraction = ( C2 × V2 )/(2×A×εo)
Answered by Expert
26th September 2018,
12:43 PM
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