Question

In: Physics

please answer all qustions and all parts and make sure you have the right answers. 1....

please answer all qustions and all parts and make sure you have the right answers.

1. An electric field is given by Ex = 3.5x3 kN/C. Find the potential difference between the points on the x axis at x = 1 m and x = 3 m.

..... kV

---------------------------------

2. (a) Find the maximum net charge that can be placed on a spherical conductor of radius 12 cm before dielectric breakdown of the air occurs.
.... µC

(b) What is the potential of the sphere when it carries this maximum charge?
..... kV

-------------------------------

3. Consider an electron and a proton that are initially at rest and are separated by 1.20 nm. Neglecting any motion of the much more massive proton.

(a) What is the minimum kinetic energy with which the electron must be projected so it reaches a point a distance of 26.0 nm from the proton? Assume the electron's velocity is directed radially away from the proton.
..... J

(b) What is the electron's speed at this energy?
..... m/s

Solutions

Expert Solution

1. An electric field is given by Ex = 3.5 x3 kN/C.

The potential difference between the points on the x axis at, x = 1 m and x = 3 m which will be given as :

We know that,    Ex = - dV / dx

using an equation, dV = - Ex dx                                                      { eq.1 }

inserting the value of Ex in above eq.

dV = - (3.5 x3 kN/C) dx

integrating an above eq.

V2 - V1 = - (3.5 kN/C) x3 dx

V2 - V1 = - (3.5 kN/C) [x4 / 4]31

V2 - V1 = - [(3.5 kN/C) / 4] [(3 m)4 - (1)4]

V2 - V1 = - [(3.5 kN/C) / 4] (80 m4)

V2 - V1 = - 70 kV

2. (a) The maximum net charge which is given as :

using an equation, Ebreakdown = k Qmax / R2

Or    Qmax = Ebreakdown R2 / k { eq.2 }

where, Ebreakdown = dielectric strength of a spherical conductor = 3 x 106 V/m

k = constant = 9 x 109 Nm2/C2

R = radius of spherical conductor = 12 cm = 0.12 m

inserting these values in eq.2,

Qmax = (3 x 106 V/m) (0.12 m)2 / (9 x 109 Nm2/C2)

Qmax = 0.0048 x 10-3 C

Qmax = 4.8 x 10-6 C

Qmax = 4.8 C

(b) The potential of the sphere when it carries this maximum charge which is given as :

Vmax = k Qmax / R                                                                 { eq.3 }

inserting the values in eq.3,

Vmax = (9 x 109 Nm2/C2) (4.8 x 10-6 C) / (0.12 m)

Vmax = 360 x 103 V

Vmax = 360 kV


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