What Happens to Capacitance and Charge in Isolated Parallel Plates?

AI Thread Summary
In the discussion about isolated parallel plates, key points include the relationships between capacitance (C), charge (Q), and stored energy (U). It is noted that doubling the distance between the plates does not increase capacitance, contrary to the initial assumption. Inserting a dielectric increases both charge and capacitance, while it decreases stored energy. The electric field is inversely related to the distance, meaning increasing the distance actually decreases the electric field strength. The participant seeks clarification on their calculations and the accuracy of their true/false assessments regarding these principles.
gal_racer
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which statements are true for two oppositely charged, isolated parallel plates: C=capacitance, U=stored energy, (Q and -Q =charged on the plates). Note: isolated plates can not lose their charge.


a. when the distance is doubled, U increase.
b. inserting a dielectric increases Q
c. when the distance is halved, Q stays the same.
d. when the distance is doubled, C increases
e. insertng a dielectric increases C
f. increasing the distance increases the Electric Field
g. inserting a dielectric decreases U


by using these equations: C=Eo (a/d) , Q=CV, and U=1/2 QV, Q^2/2C
i came up with TFTTTFT, but its not right anyone has any idea of what i might of done wrong?
 
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Look at d carefully.
 
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