Calculating the Capacitance Problems and Solutions

 Problem #1

A parallel-plate capacitor has circular plates of 8.20 cm radius and 1.30 mm separation. (a) Calculate the capacitance.
(b) Find the charge for a potential difference of 120 V.

Answer;
Known:
Radius, R = 8.2 x 10^─2 m,
Then area, A = πR² = π(8.2 x 10^─2 m)² = 0.021 m²
d = 1.30 mm = 1.30 x 10^─3 m
potential difference, V = 120 V

(a) capacitance given by

C = ϵ₀A/d
   = (8.85 x 10^─12 F/m)(0.021 m²)/(1.30 x 10^─3 m)
C = 1.44 x 10^─10 F = 144  pF

(b) the charge for a potential difference of 120 V, given by

V = q/C

q = CV = (1.45 x 10^─10 F)(120 V) = 1.72 x 10^─8 C = 17.3 nC

Problem #2
The plates of a spherical capacitor have radius 38.0 mm and 40.0 mm. (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance?

Answer;
Known:
spherical capacitor radius, b = 38.0 mm = 38.0 x 10^─3 m and a = 38.0 mm = 40.0 x 10^─3 m, then

(a) the capacitance given by

C = 4πϵ₀ab/(b ─ a)

   = (8.85 x 10^─12 F/m)(40.0 mm)(38.0 mm)/(40.0 mm ─ 38.0 mm)

C = 8.45 x 10^─11 F = 84.5 pF

(b) LeT the area required be A. Then

C = 4πϵ₀A/(b ─ a)

A = C(b ─ a)/4πϵ₀

   = (8.45 x 10^─11 F)(40 mm ─ 38.0 mm)/(4π x 8.85 x 10^─12 F/m)

A = 191 x 10^─4 m² = 191 cm²

Problem #3
What is the capacitance of a drop that results when two mercury spheres, each of radius R = 2.00 mm, merge?

Answer;
Given that a drop of mercury of radius r = 2.00 mm is an isolated sphere. When two drops each of radius r = 2.00 mm merge then the final volume of the resulting mercury drop is given by

V = 2 x (4πr³/3) = 2 x 4π x (2.0 x 10^─3 m)³/3 = 1.00 x 10^─4 m³

Let the final radius be R

4πR³/3 = 2 x 4π x (2.0 x 10^─3 m)³/3

R = 2.52 x 10^─3 m

The capacitance of a drop is given by
C = 4πϵ₀R = 4π x 8.85 x 10^─12 F/m x 2.52 x 10^─3 m

C = 2.799 x 10^─13 F = 0.28 pF

Problem #4
You have two flat metal plates, each of area 1.00 m², with which to construct a parallel-plate capacitor. (a) If the capacitance of the device is to be 1.00 F, what must be the separation between the plates?

Answer;
Known:
Area, A = 1.00 m²
Capacitance, C = 1.00 F

the separation distance between two plates, given by

C = ϵ₀A/d

d = ϵ₀A/C = (8.85 x 10^─12 F/m)(1.00 m²)/(1.00 F) = 8.85 x 10^─12 m = 8.85 pm

Problem #5
If an uncharged parallel-plate capacitor (capacitance C) is connected to a battery, one plate becomes negatively charged as electrons move to the plate face (area A). In Fig. 25-26, the depth d from which the electrons come in the plate in a particular capacitor is plotted against a range of values for the potential difference V of the battery.The density of conduction electrons in the copper plates is 8.49 x 10²⁸ electrons/m³. The vertical scale is set by d_s = 1.00 p_m, and the horizontal scale is set by V_s = 20.0 V.What is the ratio C/A?

Answer;
The charge on the surface of the plate for a given potential is,

q = Ne = (nAd)e (*)

Here, d is the density of the conducti electrons, A is the area, e is the charge of the electron
The collected charge by the plate is,

q = CV   (**)

use the euation on (*) and (**) to solve for the ratio of C/A

CV = (nAd)e

C/A = nde/V
   
        = (8.49 x 10²⁸ electrons/m³)(1.00 x 10^─12 m)(1.6 x 10^─19 C)/20 V =

C/A = 6.792 x 10^─4 F/m²  

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