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Answer :
Decreasing d decreases C, Decreasing d does not decrease Q, Inserting a dielectric with κ increases U
The statements can be evaluated one by one:
- True: According to the formula for capacitance of a parallel plate capacitor, decreasing the plate separation (d) increases the capacitance (C).
- False: The charge (Q) on the capacitor is determined by the voltage (V) and capacitance (C), not the plate separation (d).
- True: Inserting a dielectric with a dielectric constant (κ) will increase the stored energy (U) of the capacitor.
- False: Increasing the plate separation (d) decreases the capacitance (C) and thus decreases the stored energy (U) of the capacitor.
- False: Inserting a dielectric after disconnecting the capacitor from the battery does not affect the capacitance (C).
- False: Inserting a dielectric after disconnecting the capacitor from a battery does not affect the voltage (V).
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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