Question

Suppose we have a piece of metal sheet of length l = 5.8 cm, and width w = 10.2 cm to make a capacitor.

(a) If we cut it in two equal halves and make it into a parallel capacitor with plate separation d = 0.4 cm, find the capacitance in pF.
=____ pF

(b) If we cut it into two sections with widths 2?a and 2?b instead (a < b), and roll them up to form a cylindrical capacitor with two shells separated by the same distance d radially, find a and b (in cm).
a =___ cm

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Suppose we have a piece of metal sheet of length l = 4.2 cm, and width w = 9.90 cm to make a capacitor.

(a) If we cut it in two equal halves and make it into a parallel capacitor with plate separation d = 0.4 cm, find the capacitance in pF.
__________ pF

(b) If we cut it into two sections with widths 2?a and 2?b instead (a < b), and roll them up to form a cylindrical capacitor with two shells separated by the same distance d radially, find a andb (in cm).
a = ____________cm
b = ____________ cm

(c) Find the capacitance of this cylindrical capacitor in pF.
___________pF

a). Capacitance = A*epsilon/d = (4.2*4.95*10^-4)(8.85*10^-12)/(0.4*10^-2) = 4.6pF

b). 2*pi*r1 = 2a => a = pi*r1

2*pi*r2 = 2b => b = pi*r2

r2 - r1 = separation = 0.4cm..................................(1)

a+b = width/2 = 4.95cm

pi*(r1+r2) = 4.95 cms => r1+r2 = 1.57.................(2)

Solving (1) and (2) we get

r2 = 0.99cm & r1 = 0.59cm

Therefore,

a = pi*r1 = 1.84cms

b = pi*r2 = 3.11cms

c). Capacitance of cylindrical capacitor = 2*pi*epsilon*L/(ln(r2/r1) = 1.21pF