The principle of superposition of electric fields
In accordance with your variant to solve one of the following problems listed below. The number of problem statement and all necessary input data are reduced in the Table 1.1. If there is letter index in the number of the problem statement, then you should only answer the question, which corresponds to this index.
1 The electrostatic field is created with two endless parallel plates, which are charged uniformly with surface densities σ1 and σ2. Find force, which acts in this field on the point charge Q1 , if it’s situated:
а) between the planes;
b) outside of planes.
2 The point charge Q1 is situated in the centre of uniformly charged sphere with a radius R. Find the electrostatic intensity in two points, which lies from the centre at a distance of r1 and r2, if:
а) the charge of sphere equals Q2 ;
b) surface density of the sphere charge equals σ2.
3 The long thread, uniformly charged with the linear density τ1, is situated on an axis of the long cylinder, which has the radius of R. The cylinder is uniformly charged with the linear density of τ2. Define the electrostatic field intensity in the other points:
1) at a distance of r1 from the thread;
2) at a distance of l from the surface of cylinder.
4Two long parallel threads are uniformly charged with the linear densities of τ1 and τ2. The distance between the thread is l. Define the electrostatic field intensity at the point, which is situated at a distance of r1 from the first thread and r2 from the second thread.
5 The electrostatic field is created with the uniformly charged endless plate and sphere. The surface charge density of the plate is σ1. The radius of sphere is R, the surface charge density of the sphere is σ2. The sphere centre is situated at a distance of l from the plane. Find the electric intensity at the point, which is situated between the sphere and the plate at a distance of r1 from the plate.
TABLE OF TASK VARIANTS
Table 1.1
| Variant | Problem statement number | Q1 , nC | Q2, nC | τ1, nC /m | τ2, nC /m | σ1, nC /m2 | σ2, nC /m2 | R , cm | l , cm | r1, cm | r2, cm |
| 1а | +0.2 | – | – | – | +2 | +4 | – | – | – | – | |
| 2а | –3 | +6 | – | – | – | – | – | ||||
| – | – | –6 | –9 | – | – | – | |||||
| – | – | – | – | +10 | +40 | – | |||||
| – | – | –2 | +3 | – | – | – | |||||
| 1б | +0.2 | – | – | – | –2 | –3 | – | – | – | – | |
| 2б | +0.2 | – | – | – | – | +3 | – | ||||
| – | – | +3 | +4 | – | – | – | |||||
| – | – | – | – | –4 | –9 | – | |||||
| 1а | +0.1 | – | – | – | +2 | –3 | – | – | – | – | |
| 2б | +0.1 | – | – | – | – | –10 | – | ||||
| – | – | +1 | +2 | – | – | – | |||||
| 1б | +0.4 | – | – | – | –5 | +3 | – | – | – | – | |
| 2а | +1 | +5 | – | – | – | – | – | ||||
| – | – | +10 | –5 | – | – | – | |||||
| – | – | – | – | –6 | +8 | – | |||||
| – | – | –40 | –30 | – | – | – | |||||
| 1а | +0.3 | – | – | – | –4 | –1 | – | – | – | – | |
| 2а | –2 | –4 | – | – | – | – | – | ||||
| – | – | +8 | +4 | – | – | – | |||||
| – | – | – | – | +4 | –6 | – | |||||
| – | – | –3 | –4 | – | – | – | |||||
| 1б | +0.1 | – | – | – | +3 | +1 | – | – | – | – | |
| 2б | –0.4 | – | – | – | – | –8 | – | ||||
| – | – | –8 | +6 | – | – | – | |||||
| 1а | –0.4 | – | – | – | –2 | +4 | – | – | – | – | |
| – | – | +2 | –4 | – | – | ||||||
| 2б | –0.5 | – | – | – | – | +9 | – | ||||
| – | – | – | – | –3 | –8 | – | |||||
| – | – | +4 | +6 | – | – | – |
Problem 1.5.
DIRECT CURRENT
MAIN CONCEPTS
Ohm's law(simple circuits):
a) for homogeneous subcircuit:
,
where I – a current through the subcircuit; U – voltage on the subcircuit connection terminal; R – the resistance of the subcircuit;
Voltage on homogeneous subcircuit:
U = IR .
b) for subcircuit containing EMF source:
,
where Dj2,1= j2–j1 = U – potential difference (or voltage) on subcircuit connection terminal, e – value of Electro Motive Force (EMF) of source, which contains in the subcircuit; r – internal source resistance;
Voltage on branch of circuit containing EMF source:
U = e – IR–Ir .
c) For the closed circuit (when it is possible to reduce to one equivalent source and one resistor):
,
where R – external resistance of circuit, r – internal resistance of source.
Kirchoff’s rules (branched circuits):
For a branched circuits solution use the two Kirchoff’s rules (see example 3):
1st – junction rule: 
.
where 
– algebraic sum of a current into the junction (which is positive when a current flows in the junction, and negative when a current flows out of the junction);
2nd – closed loop rule: 
,
– algebraic sum of voltage drop on external resistors around any closed loop, and 
– algebraic sum of the voltage drop on internal resistance of sources (which are positive, when the direction of a current coincides with chosen one in advance direction of path-tracing);
– algebraic sum of the source electromotive force of the closed loop (which are positive, when the direction of extraneous force work (from – to + inside the source) coincides with chosen one in advance direction of path-tracing).