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FrozenT [24]
4 years ago
6

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
Physics
2 answers:
motikmotik4 years ago
5 0

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

<u>Now the total charge on the parallel plate capacitor is given as:</u>

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}

AVprozaik [17]4 years ago
3 0

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

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