Question

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 7.20 mm, and the length of each cylinder is 13.0 cm.

(a) What is the capacitance?
(b) What applied potential difference is necessary to produce these charges on the cylinders?

Answer 1

A)2.811 × 10^-12 F

B)4.09V

Step-by-step explanation:

(a) What is the capacitance?

Capacitance of coaxial cylinder can be determined using below expresion

C= (2π ε0 L)/ Ln[ rb/ra]

Where

ε0= permittivity of free space= 8.85×10^−12 Fm^-1

L= length of each cylinder =13.0 cm= 13×10^-2m

rb= radius of outer cylinder= 7.20 mm= 7.20×10^-3m

ra=radius of inner cylinder=0.550 mm= 0.550×10^-3 m

If we substitute the values we have,

C= (2π ε0 L)/ Ln[ rb/ra]

C= ( 2 × π × 8.85×10^−12 ×13×10^-2) / Ln[

7.20×10^-3/0.550×10^-3]

C=( 7.2288×10^-12 )/2.5729

C=2.811 × 10^-12 F

B) (b) What applied potential difference is necessary to produce these charges on the cylinders?

Vba= Q/C

Where Vba=potential difference

Q= charge on each = 11.5 pC. = 11.5×10^-12

C= capacitance= 2.811 × 10^-12 F

If we substitute the values we have

Vba=(11.5×10^-12)/2.811 × 10^-12

= 4.09V