Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and
1 answer:
Answer:-q
Explanation:
Given
Capacitor is charged to a battery and capacitor acquired a charge of q i.e.
+q on Positive Plate and -q on negative Plate.
If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by
where A=area of capacitor plate
d=Separation between plates
This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q
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And so, this is the force that the worker must apply to the crate.
Answer:
Explanation:
this is what i know so far ( some might be incorrect ) , but hope some of this helps!
