|
|
Electric Fields Questions Charge on an electron = -1·6×10-19C
Mass of an electron = 9·1×10-31kg
Question 1
| a) |
Define |
| |
|
i) |
electric field strength |
| |
|
ii) |
electric potential at a point in a field. |
| b) |
Calculate the magnitude of the electric field strength at a point 5×10-8m from an electron. |
| c) |
Calculate the magnitude of the electric potential at a point 5×10-8m from an electron. |
Question 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. |
Question 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. |
Question 4
| The diagram below shows two small isolated metal spheres having charges, Q1 = -1·5×10-12C and Q2 = +2·0×10-12C |
 |
| 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). |
Question 5
| Calculate the speed of an electron which has "fallen" freely through a p.d. of 500V. |
Question 6
| Calculate the strength of the electric field needed to accelerate alpha particles (of mass 6·68×10-27kg and charge 3·2×10-19C) from rest to a speed of 2·5×105ms-1 in a distance of 5mm. |
Question
7
|
Two positive changes are 15cm
apart as shown below.
On the line joining the
centres of the two charges there exists a
point 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.
|
Question
8
|
A small
positively charged ball is released from the
position shown in the diagram below. The mass
of the ball is m = 0.5g
and it carries a charge of magnitude Q =
5µC.
The two metal plates are charged and the
potential difference between them is V =
5V.
Find the
direction in which the ball will move when
released. (The apparatus is on the earth.) |
Question
9
| A beam of electrons enters the uniform electric field between two parallel charged plates, as shown in the diagram below. |
 |
The fluorescent screen is placed very close to the end of the plates.
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. |
|
© David Hoult 2008 |