A 4.0 μF parallel plate capacitor is connected to a 5 V battery. The work done by an external agent to slowly increase the separation between the plates to double of its original value is
Answers
Answer:
: 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
Explanation:
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Answer:
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
Explanation:
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