A potential difference of 500 V is applied across the two collinear conducting c
ID: 1589533 • Letter: A
Question
A potential difference of 500 V is applied across the two collinear conducting cylinders shown in the figure. The radius of the outer cylinder is 22.55 cm, the radius of the inner cylinder is 17.82 cm, and the length of the two cylinders is 58.7 cm. How much charge is applied to each of the cylinders? What is the magnitude of the electric field between the two cylinders?
a) The charge applied to each of the cylinders = q = ___ nC
b) The magnitude of the electric field just outside the inner surface = E1 = ____ V/m
c) The magnitude of the electric field just inside the outter surface = E2 = ____ V/m
Explanation / Answer
a) C = 2*pi*epsilon*L/ln(b/a)
= 2*pi*8.854*10^-12*0.587/ln(0.2255/0.1782)
= 1.387*10^-10 F
The charge applied to each of the cylinders = q = C*V
= 1.387*10^-10*500
= 69.35*10^-9 C or 69.35 nc
b) linear charge density on inner cyllinder.
lamda = q/L
= 69.35*10^-9/0.587
= 1.18*10^-7 C/m
so, the electric field just outside the inner surface = E1 = 2*k*lamda/a
= 2*9*10^9*1.18*10^-7/0.1782
= 1.19*10^4 V/m
c) the electric field just inside the outer surface = E1 = 2*k*lamda/b
= 2*9*10^9*1.18*10^-7/0.2255
= 9.42*10^3 V/m
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