CapacitanceHard
Question
The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
Options
A.the energy stored in the capacitor will become K-times
B.the electric field inside the capacitor will decrease to K-times
C.the force of attraction between the plates will increase to K2 − times
D.the charge on the capacitor will increase to K-times
Solution
Battery connected V = constant
U′ =
KCV2 = KU ⇒ Increase by K−times
E =
= constant
F =
⇒ F = 
⇒ F′ = 
= K2F
Þ Increase by K2−times
Q = CV ⇒ Q′ = KCV = KQ ⇒ Increase by K−times.
U′ =
KCV2 = KU ⇒ Increase by K−times E =
= constant F =
⇒ F = ⇒ F′ = = K2FÞ Increase by K2−times
Q = CV ⇒ Q′ = KCV = KQ ⇒ Increase by K−times.
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