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A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. After being disconnected from the battery, decreasing d decreases C. With the capacitor connected to the battery, decreasing d decreases Q. With the capacitor connected to the battery, inserting a dielectric with κ will increase U. With the capacitor connected to the battery, increasing d increases U. After being disconnected from the battery, inserting a dielectric with κ will decrease C. After being disconnected from the battery, inserting a dielectric with κ will increase V.

Respuesta :

Answer:

true    C

false   a, b, d, e

Explanation:

Capacitance in a condensate can be found with any of the following equations

       C = Q / DV

       C = e or A / d

with these two expressions we answer the final statements

a) False. From the equations above we see that by decreasing the distance between the plates (d) the capacitance increases, by disconnecting the ideal capacitor the charge remains constant

b) False. After disconnecting the battery the charge on the plates remains constant

The energy stored in a capacitor is given by

     U = k q / DV = k e A / d

c) True. From the previous equation we see that the energy is proportional to the dielectric

d) False. Capacitance increases with dielectric

e) False. The dielectric creates a field that opposes the field of the capacitor, whereby the total electric field decreases accordingly as the field and the voltage are proportional the potential difference must decrease

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