How many excess electrons must be added to one plate and removed from the other

Question

How many excess electrons must be added to one plate and removed from the other to give a 5nF parallel plate capacitor 30 mJ stored energy? How could you modify the geometry of this capacitor to get it store 60 mJ of energy without changing the charge on its plates? I am very interested in how to manipulate one of the formulas for potential energy to derive modifying the geometry? How many excess electrons must be added to one plate and removed from the other to give a 5nF parallel plate capacitor 30 mJ stored energy? How could you modify the geometry of this capacitor to get it store 60 mJ of energy without changing the charge on its plates? I am very interested in how to manipulate one of the formulas for potential energy to derive modifying the geometry?

Explanation / Answer

The capacitance of the parallel plate capacitor is C = 5 nF = 5 * 10-9 F The energy stored in the capacitor is E = 30 mJ = 30 * 10-3 J We know from the relation E = (1/2)CV2 where V is the potential difference across the parallel plate capacitor or V2 = (2E/C) or V = (2E/C)1/2 The charge on the plates of the capacitor is Q = C * V We know that Q = n * e or n * e = C * V or n = (C * V/e) where n is the excess electrons that must be added to one plate and removed from the other to give a 5nF parallel plate capacitor 30 mJ stored energy and e = 1.6 * 10-19 C When the energy stored is U = 60 mJ = 60 * 10-3 J then (1/2)CV2 = U or (1/2) * (Q/V) * V2 = U or (1/2) * Q * V = U or V = (2U/Q) or E * d = (2U/Q) where E is the electric field between the plates of the capacitor and d is the separation between the plates. or d = (2U/E * Q) In the above equation U,E and Q are constant.Therefore,the separation between the plates should be doubled to store 60 mJ of energy without changing the charge on its plates. The charge on the plates of the capacitor is Q = C * V We know that Q = n * e or n * e = C * V or n = (C * V/e) where n is the excess electrons that must be added to one plate and removed from the other to give a 5nF parallel plate capacitor 30 mJ stored energy and e = 1.6 * 10-19 C When the energy stored is U = 60 mJ = 60 * 10-3 J then (1/2)CV2 = U or (1/2) * (Q/V) * V2 = U or (1/2) * Q * V = U or V = (2U/Q) or E * d = (2U/Q) where E is the electric field between the plates of the capacitor and d is the separation between the plates. or d = (2U/E * Q) In the above equation U,E and Q are constant.Therefore,the separation between the plates should be doubled to store 60 mJ of energy without changing the charge on its plates.

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