How does a dielectric affect the capacitance of a capacitor?

[catch22]

Homework Statement

Homework Equations

The Attempt at a Solution

for #14, I remember a battery maintains the potential therefore, V should be constant; dielectric increases charge on plates and increases capacitance but I couldn't find an option that matched.

The answer key says "d" but I believe V should stay the same so perhaps there is a typo?
Can anyone confirm this?

 

[catch22]

Another question:

Could someone explain this to me? Answer key says C.
I thought in series, Ceq = C1 + C2 + C3...

and in parallel Ceq = 1/C1 + 1/C2 + 1/C3...

 

[honlin]

For the first part, i agree with you that the potential difference between two plates should be the same because it is still attached to a battery. For the second question, think it this way: which quantity is the same for all the capacitors when you connect them in parallel, while which quantity is the same when you connect them in series?

 

[catch22]

For the first part, i agree with you that the potential difference between two plates should be the same because it is still attached to a battery. For the second question, think it this way: which quantity is the same for all the capacitors when you connect them in parallel, while which quantity is the same when you connect them in series?

in series, Q is constant through out.

Parallel , V is constant through out.

 

[ehild]

Could someone explain this to me? Answer key says C.
I thought in series, Ceq = C1 + C2 + C3...

and in parallel Ceq = 1/C1 + 1/C2 + 1/C3...

Your formulae for the equivalent capacitors are wrong. They are not the same as in case of resistors!

 

[honlin]

in series, Q is constant through out.

Parallel , V is constant through out.

Yea, you can work out from there. Try to find the total energy from the Q,C,V.

 

[catch22]

Your formulae for the equivalent capacitors are wrong. They are not the same as in case of resistors!

whoops, for Ceq = C1 + C2 + C3... in parallel

Ceq = 1/C1 + 1/C2 + 1/C3... in series