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ELECTRIC FIELDS
     
    Charge on an electron = -1.10-19C
    Mass of an electron = 9.10-31kg
     
1. a) Define
     i)  electric field strength
    ii) electric potential at a point in an electric field. 
  b) Calculate the magnitude of the electric field strength at a point 10-8m from an electron.
  c) Calculate the magnitude of the electric potential at a point 10-8m from an electron.
     
2.   Sketch graphs (with distance on the horizontal axis) of the following
  a) field strength against distance from a point charge
b) potential against distance from a point charge
  c) field strength against distance from the centre of a charged metal sphere of radius R (in this case the distance axis should go from zero to a value much greater than R)
  d) potential against distance from the centre of a charged metal sphere of radius R (again the distance axis should go from zero to a value much greater than R)
For the next two graphs, refer to the diagram below
     
   
     
e) field strength at point p against x
  f) potential at point p against x.
     
3.   Calculate the amount of electrical potential energy gained by an electron when it is moved from infinity* to a point 0.5cm from another electron.
    *ok not like, literally, infinity; just a long way off, which for an electron might be a couple of metres!
     
4. The diagram below shows two small isolated, charged metal spheres.
     
   
Q1=-1.10-12C   Q2=+2.10-12C
   
     
  a) Calculate the magnitude of the potential at point p1
  b) Find the magnitude and direction of the field strength at point p2
    Assume that the spheres are in a vacuum (or air).
     
5.   Calculate the speed of an electron which has "fallen" freely through a potential difference of 500V.
     
6.   Calculate the strength of the electric field needed to accelerate alpha particles (of mass 6.610-27kg and charge 3.10-19C) from rest to a speed of 2.105ms-1 in a distance of 5mm.
     
7.   Two positive charges are 15cm apart as shown below.
     
   
     
    On the line joining the centres of the two charges there exists a point, p at which the resultant field strength is zero (a “neutral point”).
  a) Explain why this neutral point exists.
  b) Calculate the distance, r of this point from the charge Q1.
8.   A small positively charged ball is released from the position shown in the diagram below.
     
   
     
    The mass of the ball is 0.5g and it carries a charge of magnitude 4.9µC.
    The two metal plates are charged and the potential difference between them is 5V.
    Find the direction in which the ball will move when released.
    The apparatus is at a place where the acceleration due to gravity is 9.8ms-2
     
9.   A beam of electrons enters the uniform electric field between two parallel charged plates, as shown in the diagram below.
     
   
     
    The voltage across the plates is 1.15V and the electrons enter the field with a velocity of 106ms-1, parallel to the plates.
  a) At what distance from O will the electrons hit the screen?
  b) Calculate the magnitude and direction of the velocity of the electrons at the instant when they hit the screen.
  c) Calculate the kinetic energy possessed by one electron at the instant when it hits the screen.
    Answer in Joules then convert to electron volts.
     
10.    An electron of charge, e, and mass, m, enters the uniform electric field between two parallel charged plates, as shown in the diagram below.
     
   
     
    The potential difference across the plates is V and the distance between them is d.
    The initial velocity of the electron is of magnitude, u and is parallel to the plates.
    The point of entry into the field is midway between the two plates.
The electron hits the lower plate after having moved a distance x in the field (x measured parallel to the plates).
    Show that the distance x is given by
   
    Assume that the field starts abruptly at the end of the plates and that it is uniform between the plates.
    You will have noticed that we are ignoring the force of gravity on the electron.
    If you want to see why, try calculating the ratio of the force of gravity on the electron to the electric force for a potential difference of, say, 1V... that should convince you that we can certainly ignore the force of gravity!
     
     
 
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