An initially uncharged air-filled capacitor is connected to a 5.67-V charging source. As a result, 3.49 × 10-5 C of charge is transfered from one of the capacitor\'s plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant of this substance is 5.99. Find the capacitor\'s potential difference and charge after the insertion.

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Vespertilio Vespertilio · Answer 1 · 2019-11-23T14:46:44+01:00

Answer:

V = 5.67 V

Q' = 2.09 x 10^-4 C

Explanation:

Potential difference, V = 5.67 V

Charge, Q = 3.49 x 10^-5 C

dielectric constant, K = 5.99

Capacitance of the capacitor

C = Q / V = 3.49 x 10^-5 / 5.67 = 6.15 x 10^-6 F

Now the dielectric is inserted, so,

C' = K x C = 5.99 x 6.15 x 10^-6 = 3.687 x 10^-5 F

After insertion of dielectric, as the battery is connected, so potential difference remains same.

Charge, Q' = C' x V = 3.687 x 10^-5 x 5.67 = 2.09 x 10^-4 C

andromache andromache · Answer 2 · 2022-02-15T14:15:45+01:00

The capacitor\'s potential difference and charge after the insertion is V = 5.67 V and Q' = 2.09 x 10^-4 C

The capacitance of the capacitor is

C = Q / V

= 3.49 x 10^-5 / 5.67

= 6.15 x 10^-6 F

Now the dielectric is inserted

So,

C' = K x C

= 5.99 x 6.15 x 10^-6

= 3.687 x 10^-5 F

Now

Charge, Q'

= C' x V

= 3.687 x 10^-5 x 5.67

= 2.09 x 10^-4 C

learn more about the substance here: https://brainly.com/question/149989