Question

A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
25.0

c
m
2
. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.

(a) Determine the charge on the plates before and after immersion.

(b) Determine the capacitance and potential difference after immersion.

(c) Determine the change in energy of the capacitor.

Answer 1

Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

$C = \frac{\epsilon _0A}{d}$

$C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}$

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

$U = \frac{q^{2}}{2C}$

$U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J$

Final energy

$U' = \frac{q^{2}}{2C'}$

$U' = \frac{(3.2\times 10^{-10})^{2}}{2\times 100.64\times 10^{-12}}=5.08\times 10^{-10}J$

Answer 2

The charge on the plates before and after the immersion is $3.21 \times 10^{-10}\;C$.

Given the following data:

• Plate separation, d = 1.76 cm to m = 0.0176 m.

• Plate area, A = 25 $cm^2$ to $m^2$ = 0.0025 $m^2$.

• Potential difference, V = 255 Volts.

Scientific data:

• Permittivity of free space = $8.854 \times 10^{-12}$

• Dielectric constant of water = 80

Note: We would assume that the liquid is an insulator.

a. To determine the charge on the plates before and after immersion:

First of all, we would calculate the capacitance of this capacitor by using this formula:

$C = \frac{\epsilon_o A }{d} \\\\C= \frac{8.854 \times 10^{-12} \times 0.0025}{0.0176} \\\\C= \frac{2.2135 \times 10^{-14} }{0.0176}\\\\C=1.258 \times 10^{-12}\;Farad$

Now, we can determine the charge:

$q = CV\\\\q = 1.258 \times 10^{-12} \times 255\\\\q=3.21 \times 10^{-10}\;C$

Note: The charge on the plates is the same before and after immersion because the battery is disconnected.

b. To determine the capacitance and potential difference after immersion:

Before:

Capacitance, C = $1.258 \times 10^{-12} \;Farad$

After:

$Capacitance = kC\\\\C'= 80 \times 1.258 \times 10^{-12}\\ \\C'= 100.64 \times 10^{-12}\; F$

For the potential difference:

Before:

Potential difference, V = 255 Volts.

After:

$Potential \;difference = \frac{V}{k} \\\\Potential \;difference = \frac{255}{8}$

Potential difference, V =3.19 Volts.

c. To determine the change in energy of the capacitor:

$\Delta E= U'-U\\\\\Delta E=\frac{q^2}{2C'} - \frac{q^2}{2C}\\\\\Delta E=\frac{(3.21 \times 10^{-10})^2}{2\times 100.64 \times 10^{-12}} -\frac{(3.21 \times 10^{-10})^2}{2\times 1.258 \times 10^{-12}} \\\\\Delta E= 4.07 \times 10^{-8} -5.08 \times 10^{-10} \\\\\Delta E= 4.02 \times 10^{-8}\;Joules$

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