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

Answer:

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

Answer 2

Answer:

D. 5.2 x 10^5

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:

$Q=V.A\frac{\epsilon_o}{d}$

where:

$A=$ area of each plate

$\epsilon_0=$ permittivity of free space

$Q=1.5\times (0.0025^2\times 8.85\times 10^{-12}\times\frac{1}{10^{-3}} )$

$Q=8.3\times 10^{-14}\ C$

Now the no. of electrons:

$n=\frac{Q}{e}$

where:

$e=1.6\times 10^{-19}\ C$ is the charge on an electron

$n=\frac{8.3\times 10^{-14}}{1.6\times 10^{-19}}$

$n\approx5.2\times 10^{5}$