One of the common uses for a coaxial cable is to transmit radio frequencies. One

Question

One of the common uses for a coaxial cable is to transmit radio frequencies. One such coaxial cable of length 45.0 m consists of an inner conducting cylinder with a radius of 1.15 mm and an outer conducting cylinder whose inner radius is 3.25 mm. The charges on the inner and outer cylinders are 7.10 µC and +7.10 µC, respectively. (a) Determine the potential difference between the two cylinders along a path from the inner cylinder to the inside edge of the outer cylinder. Use the fact that the magnitude of the electric field at a distance r from the axis of the inner cylinder is given by E = 2k r where k is Coulomb's constant and is the linear charge density. V (b) If the space between the two cylinders is filled with air, determine the capacitance of the cable. nF

Explanation / Answer

Here ,

inner radius , a = 1.15 mm

outer radius , b = 3.25 mm

electric field inside the conductors , E = 2 * k * lamda/r

E = 2 * 9*10^9 * (7.1 *10^-6/45)/r

E = 2840/r

Now , potential difference , V = integration(E.dr) from 3.25 mm to 1.15 mm

V = integration(2840/r dr) from 3.25 mm to 1.15 mm

V = 2840 * ln(3.25/1.15)

V = 2950.5 V

the potential difference between the two cyclinders is 2950.5 V

b)

Here ,

as Q = C * V

7.1 *10^-6 = 2950.5 * C

C = 2.406 *10^-9 F

the capaitance is 2.406 nF

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.