Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca
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Attached is the solution to the above question.
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500××
=2×
=2 pF