THE UNIVERSITY OF BRITISH COLUMBIA 
 
Science 1  
Physics Assignment # 4: 
 
 ELECTRIC CHARGE  
 
 26 Jan. 2000 - finish by 02 Feb. 2000 
 
Tipler Ch. 18, problems  25, 33, 34 &  42;
- 
 
- 1.
- Units:  
Moving a charge of one 1 C [one coulomb] 
through a potential of 1 V [one volt] 
takes 1 J [one joule] of work 
(or will produce 1 J of kinetic energy).  
This handy relationship is a little clumsy in modern physics, 
which deals with elementary particles 
whose charge is quantized in units of 
 C.  
A convenient energy unit in accelerator physics 
is therefore the electron volt [eV], 
defined as the (kinetic) energy gained by an electron 
(or any other particle with the same size charge) 
when it is accelerated through a one volt [1 V] 
electric potential.1
  
So  . . .  
how many electron volts in one kilowatt-hour?2 C.  
A convenient energy unit in accelerator physics 
is therefore the electron volt [eV], 
defined as the (kinetic) energy gained by an electron 
(or any other particle with the same size charge) 
when it is accelerated through a one volt [1 V] 
electric potential.1
  
So  . . .  
how many electron volts in one kilowatt-hour?2
 
 
- 2.
- Earth's Extra Electrons:  
The Earth's atmosphere is a thin electrical insulator 
which has an excess of positive charge at higher altitude 
an excess of negative charge near the Earth's surface, 
which acts like a spherical conductor.  
This charge separation gives rise to an approximately constant 
electric field pointing down through the atmosphere.  
However, above the atmosphere there is no 
electric field due to the Earth unless it has a net charge.  
Assuming that only electrostatic and gravitational forces are involved, 
calculate how many excess electrons the Earth could hold.  
(Hint: what will happen to a free electron just outside the atmosphere?)  
 
 
- 3.
- Gauss' Law for a Sheet of Charge:  
  Imagine an infinite plane sheet of electric charge 
 with 
 units of charge per unit surface area. 
 Using your own words and drawings,3
  . . . units of charge per unit surface area. 
 Using your own words and drawings,3
  . . .
 
- 
 
- (a)
-  Using only (i) the  SUPERPOSITION PRINCIPLE 
for the total electric field due to an assembly of 
electric charges, plus (ii) simple 
 SYMMETRY arguments, deduce the direction 
of the electric field due to the plane of charge.  
 
- (b)
-  Using the preceding result plus the general 
ideas of  GAUSS' LAW and simple geometry, 
deduce the dependence of the 
magnitude E of the electric field upon x, 
the perpendicular distance away from the plane.  
 
 
 
 
- 4.
- Gauss' Law for a Line of Charge:  
 Imagine an infinite straight line of electric charge 
 with 
 units of charge per unit length.  
 Using your own words and drawings,  . . . units of charge per unit length.  
 Using your own words and drawings,  . . .
 
- 
 
- (a)
-  Using only (i) the  SUPERPOSITION PRINCIPLE 
for the total electric field due to an assembly of 
electric charges, plus (ii) simple 
 SYMMETRY arguments, deduce the direction 
of the electric field due to the line of charge.  
 
- (b)
-  Using the preceding result plus the general 
ideas of  GAUSS' LAW and simple geometry 
(surface area of a cylinder, etc.), 
deduce the dependence of the 
magnitude E of the electric field upon r, 
the perpendicular distance away from the line.  
 
 
 
 
- 5.
- 
Triangle of Charges: 
 Three charges are placed at the vertices of 
 an equilateral triangle of side a (as shown) 
 and a fourth charge is placed at the centre of the triangle.  
 
 
  
 
- (a)
- Sketch the electric field lines 
inside and outside (nearby) the triangle.  
 
- (b)
- Derive an expression for the vector electric field 
   at the centre charge 
  due to the three other charges at the vertices of the triangle. at the centre charge 
  due to the three other charges at the vertices of the triangle.
 
 
 
Jess H. Brewer 
2000-01-25