Question

How many electrons are contained on the negatively-charged side of a capacitor when it is fullycharged if it is connected to a 1.5 V battery, the capacitor’s square plates are 0.25 cm x 0.25 cm, and the plates’ separation is 1 mm?

A. 1.8 x 10^9
B. 3.8 x 10^8
C. 2.5 x 10^7
D. 5.2 x 10^5
E. 7.5 x 10^4

Answer 1

D. 5.2 x 10^5

Step-by-step explanation:

Given:

voltage of the battery connected,
`V=1.5\ V`

sides of square capacitor,
`s=0.0025\m`

separation between the plates,
`d=10^(-3)\ m`

Now the total charge on the parallel plate capacitor is given as:

`Q=V.A(\epsilon_o)/(d)`

where:

`A=` area of each plate

`\epsilon_0=` permittivity of free space

`Q=1.5* (0.0025^2* 8.85* 10^(-12)*(1)/(10^(-3)) )`

`Q=8.3* 10^(-14)\ C`

Now the no. of electrons:

`n=(Q)/(e)`

where:

`e=1.6* 10^(-19)\ C` is the charge on an electron

`n=(8.3* 10^(-14))/(1.6* 10^(-19))`

`n\approx5.2* 10^(5)`

Answer 2

Step-by-step explanation:

Capacitance of the capacitor

C = ε₀ A / d

(8.85 x 10⁻¹² x .25² x 10⁻⁴) / 1 x 10⁻³

C = .553125 x 10⁻¹³ F

Charge = capacitance x volt

= .553125 x 10⁻¹³ x 1.5

= .8296875 x 10⁻¹³ C

no of electrons

= charge / charge on one electron

= .8296875 10⁻¹³/ 1.6 x 10⁻¹⁹

= 5.2 x 10^5