The work done in raising the potential of a capacitor from v volt to 220 volt is doubled then the work done in increasing the potential on the same capacitor from to be 24 we will be
Answers
The energy initially stored by the capacitor is 1/2 C*V^2 or Q^2 / 2*C
When the plate separation is doubled, the charge remains constant and the capacitance is halved. The work done in separating the plates is the difference in energy stored: this is
Q^2 / 2*0.5*C - Q^2 / 2*C This represents double the initial energy less the initial energy. The work done in adding this energy is Q^2 / 2*C
Assigning values: ( 2*10–4 )^2 / ( 2* 10 *10^-6 ) = 2 mJ
Answer:
A capacitor has an even electric field between the plates of strength E (units: force per coulomb). So the voltage is going to be E×distance between the plates. Therefore increasing the distance increases the voltage. ... As charge remains constant, per charge energy increases as well (that is potential difference).
Explanation:
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