We know energy per unit volume stored in electric field =12ε0E2 in parallel plate capacitor E=6ε0 Energy density =12ε062ε2o Energy=12ε01ε0(QA)2⋅(A⋅d)→volume E=121ε0Q2dA We know C=ε0Ad E=12Q2C(Q=CV) E=12CV2
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Q1
A capacitor of capacitance 5.00μF is charged to 24.0V and another capacitor of capacitance 6.0μF is charged to 12.0V. (a)Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors.
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Q2
A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
(a) The electric field in the capacitor
(b) The charge on the capacitor
(c) The potential difference between the plates
(d) The stored energy in the capacitor
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Q3
A capacitor of capacitance 5⋅00 µF is charged to 24⋅0 V and another capacitor of capacitance 6⋅0 µF is charged to 12⋅0 V. (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go?
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Q4
(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field. (b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.
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Q5
A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is :