Question

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the other is filled with a dielectric for which k ! 3.00;both capacitors have a plate area of 5.00 $ 10#3 m2 and a plate separation of 2.00 mm.

Answer 1

`Q=7.9* 10^(-10)\ C`

Step-by-step explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

`A=5* 10^(-3)\ m^2`

We Capacitance given as

For air

`C_1=(\varepsilon _oA)/(d)`

`C_1=(\varepsilon _oA)/(d)`

`C_1=(8.85* 10^(-12)* 5* 10^(-3))/(2* 10^(-3))`

`C_1=2.2* 10^(-11)\ F`

`C_2=(K\varepsilon _oA)/(d)`

`C_2=(3* 8.85* 10^(-12)* 5* 10^(-3))/(2* 10^(-3))\ F`

`C_2=6.6* 10^(-11)\ F`

Net capacitance

C=C₁+C₂

`C=8.8* 10^(-11)\ F`

We know that charge Q given as

Q= C V

`Q=12* 6.6* 10^(-11)\ C`

`Q=7.9* 10^(-10)\ C`