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A fully charged parallel-plate capacitor remains connected to a battery while a dielectric is slid between the plates. Do the following quantities increase, decrease, or stay the same?

Select one or more:

a. delta V increases

b. Energy stored increases

c. Q stays the same

d. Energy stored stays the same

e. Q decreases

f. delta V decreases

g. C stays the same

h. C increases

i. E increases

j. E stays the same

k. delta V stays the same

l. E decreases

m. Q increases

n. Energy stored decreases

o. C decreases

Respuesta :

Answer:

Explanation:

Let the potential difference remains same. So, by insertion the dielectric, the capacitance

C' = KCo

As V remain same, so Q' = KQo

Energy, U' = 1/2 C'V² = KCo

Electric field, E' = E/ K

Where, k be the dielectric constant.

Thus, V remains same

Energy increases

C increases

E decreases

Q increases.