Magnetics coaxil quest?

Question

Hey Guys i have tried to attempt some of these questions i just dont understand where to start from i need the main formulas that can help me do these questions thanks alot

a) coaxial line consists of an inner conductor of radius RA = 1 cm

and outer conductor with an internal radius of RB = 10cm. The

internal conductor is charged positively with a linear charge density

of ρℓ = +100 nC/m. Calculate the potential difference between RA

and RB.

(ii) Find the total effective capacitance of the coaxial line described in

(i) if the length of the line is 30 cm and the region between the

conductors is filled with dielectric with relative permittivity of 2.5.

(e) A coaxial line consisting of inner and outer conductors with unknown

radii is located in zero gravity conditions and has a potential difference of

8 kV between electrodes. A particle with a charge of +1 mC and with a

mass of 40 g is located almost on the surface of the central, positively

charged electrode.

The particle is released from its location with zero velocity initially and

starts to move towards the opposite electrode. Calculate the velocity

which the particle will have when it reaches the opposite electrode.

Answer 1

a) The lengths of the conductors are not given. Therefore, we will assume that the conductors are infinitely long (as is assumed by default for coaxial line). Therefore, only the radial component of electric field is of importance here.

Consider a cylindrical Gaussian surface of length a and radius x between the two conductors. Since only the radial component of electric field is important, we will consider only the electric field along the radius of this cylinder i.e. perpendicular to the Gaussian surface.

Let electric field at a point on the Gaussian surface = E

Area of the Gaussian surface = 2*π*x*a

E is perpendicular to the surface.

Therefore, electric flux = E*2*π*x*a

Total charge inside the volume enclosed by the Gaussian surface = a*ρℓ

By Gauss's law = electric flux through Gaussian surface = charge inside the volume inclosed by the surface/ε0

Or,

E*2*π*x*a = a*ρℓ/ε0

Divide both sides by a

E*2*π*x = ρℓ/ε0

Or, E = ρℓ/(2π*ε0*x)

Potential at Ra - potential at RB = integral ρℓ/(2π*ε0*x) dx from RA to RB

= {ρℓ/(2π*ε0)} *(lnx) from RA to RB

= {ρℓ/(2π*ε0)} *(ln RB - ln RA)

= {ρℓ/(2π*ε0)} *ln(RB/RA)

= {100*10^-9/(2*3.14*8.85 * 10^-12)} * ln(10/1)

= 4143 volts

Edward: You have used V(r) =E(r) r.

That is wrong. You should use integral. What you have written is valid only if E(r) does not depend on r. But in the given question, it depends on r.

ii) Charge Q = charge density * length = (100 nC/cm)* 30 cm

= 3000 nC = 3000 * 10^-9 C = 3*10^-6 C

Potential difference is inversely proportional to permittivity. Potential drop with no dielectric = 413 volts

Therefore, potential drop with dielectric of relative permittivity 2.5

= 413/2.5 volts = 165.2 volts

Capacitance C = Q/V = 3*10^-6/165.2 farad = 0.018 * 10^-6 farad

= 18 * 10^-3 farad = 18 mili farad

e) Electric potential difference = work done per unit charge by electric field.

Therefore, work done per unit charge by electric field = 8 kV = 8000 V

Work done by electric field on 1 mC i.e. 10^-3 C charge = 8000 * 10^-3 = 8 Joule

By work energy theorem, this work done = change in kinetic energy of the charge.

1/2 *mv^2 = 8 Joule

m = mass of charge = 40 g = 0.04 kg

v = speed of charge = ?

1/2 *0.04*v^2 = 8

Or v^2 = 8 * 2/0.04 = 400

Or, v = 20 m/s

Answer 2

a)

i)Based on Gauss Law the field enclosed by a sphere of radius r equal to the charge q contained within that sphere over permittivity of the medium ( fro free space lets call it e0)

4 pi E(r) r = q/e0  or

E(r)= q / (4 pi e0 r)  

Similarly is for a line charge of charge density g

q= gL  and a cylindrical geometry of an very long cylinder (ignoring the flat sides of the cylinder)

2 pi R E(r) L  = gL / e0

E(r)=  g / (2 pi e0 r)  

Since V(r) =E(r) r we have

V(r)=  g / (2 pi e0 )  

ii)

V= q/C

Then C = q/V

The charge Q is the product of charge density g and length of wire L

q=gL

C= gL/V

C= gL/[ g / (2 pi e0 )  ]

C=(2 pi e0 )  /L

iii) Just plug the numbers in

e) Potential energy becomes kinetic energy

Pe= qV

Ke= 0.5 m v^2 then

v= sqrt(2 q V /m)

Does that help?