Question

A homemade capacitor is assembled by placing two 10-in. pie pans 2 cm apart by a piece of glass and connecting them to the opposite terminals of a 12-V battery. Estimate (a) the capacitance, (b) the charge on each plate, (c) the electric field halfway between the plates, and (d) the work done by the battery to charge the plates.

Answer 1

Answer:

capacitance  =  2.242 × $10^{-11}$ F

charge = 1.345 × $10^{-10}$  C

electric field = 600 V/m

work done = 8.07 × $10^{-10}$ J

Explanation:

given data

battery c = 12 V

diameter = 10 in

distance d = 2 cm = 2×$10^{-2}$ m

to find out

capacitance , charge on plate, electric field and work done

solution

we know radius is diameter / 2

so radius r = 10 / 2 = 5 in = 0.127 m

and capacitance  formula that is

capacitance  = A∈/d

put here all value

capacitance  = πr² ∈/d

capacitance  = π(0.127)² ×8.85 ×$10^{-12}$ /2×$10^{-2}$

capacitance  =  2.242 × $10^{-11}$ F

and

charge on plate is express as

charge = capacitance ×  c

we know here 2 plate so on  1 plate c is 6

charge = 2.242 × $10^{-11}$ ×  6

charge = 1.345 × $10^{-10}$  C

and

electric field is express as

electric field = c / d

electric field = 12 / 2×$10^{-2}$

electric field = 600 V/m

and

work done is express as

work done = 1/2 × charge × C

put here value

work done = 1/2 × (1.345 × $10^{-10}$) (12)

work done = 8.07 × $10^{-10}$ J

Answer 2

Electric field is the ratio of potential difference between two points and the distance between two point.

• a) The value of the capacitance is $2.242\times10^{-11}$ F.

• b) The charge on each plate is $1.345\times10^{-10}$ C.

• c) The  electric field halfway between the plates is 600 V/m.

• d) The work done by the battery to charge the plates is  $8.07\times10^{-10}$ J.

What is electric field?

Electric field is the ratio of potential difference between two points and the distance between two point.

Given information-

Diameter of the ple pans is 10 in.

Thus the radius of the pie pans is 5 in or 0.127 m.

The distance between the two pans is 2 cm or 0.02 m.

• (a) The capacitance,

Capacitance can be find out using the following formula as,

$C=\dfrac{\pi r^2\times \varepsilon_0 }{d}$

Here, $\varepsilon_0$ is the permittivity ($8.85\times10^{-12}$).

Put the values as,

$C=\dfrac{\pi (0.127)^2\times8.85\times 10^{-12}}{0.02}\\C=2.242\times10^{-11}\rm F.$

Thus the value of the capacitance is $2.242\times10^{-11}$ F.

• (b) the charge on each plate,

The charge is the product of capacitance and the potential difference.

There is total voltage is 12 V and the number of plate are 2. Therefore the charge on one plate should be 6 V.

Thus the charge can be given as,

$q=2.242\times10^{-11}\times6\\q=1.345\times10^{-10}\rm C$

Thus the charge on each plate is $1.345\times10^{-10}$ C.

• (c) the electric field halfway between the plates

Electric field is the ratio of potential difference between two points and the distance between two point.

As the value of potential difference is 12 V and the distance is 0.02 m. Thus the electric field is,

$E=\dfrac{12}{0.02} \\E=600\rm V/m$

Thus the  electric field halfway between the plates is 600 V/m.

(d) the work done by the battery to charge the plates.

Work done is the half of the product of potential difference and the charge. Thus the work done by the battery is,

$W=\dfrac{1}{2}\times(12)(1.345\times10^{-10})\\W=8.07 \times10^{-10} \rm J$

Thus the work done by the battery to charge the plates is  $8.07\times10^{-10}$ J.

Hence,

• a) The value of the capacitance is $2.242\times10^{-11}$ F.

• b) The charge on each plate is $1.345\times10^{-10}$ C.

• c) The  electric field halfway between the plates is 600 V/m.

• d) The work done by the battery to charge the plates is  $8.07\times10^{-10}$ J.

Learn more about the electric field here;

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