




Charge on an electron = 1.6×10^{19}C 


Mass of an electron = 9.1×10^{31}kg 



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 5×10^{8}m
from an electron. 

c) 
Calculate the magnitude of the electric potential at a
point 5×10^{8}m
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. 





Q_{1}=1.5×10^{12}C 

Q_{2}=+2.0×10^{12}C 








a) 
Calculate the magnitude of the potential at point p_{1} 

b) 
Find the magnitude and direction of the field strength
at point p_{2} 


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 6.68×10^{27}kg
and charge 3.2×10^{19}C)
from rest to a speed of 2.5×10^{5}ms^{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 Q_{1}. 



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 10^{6}ms^{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! 





