Q. 83.7( 81 Votes )
Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?
(a) The diagram is shown as below:
AB = 20 cm
∵ O is the midpoint of the line AB.
∴ AO = OB = 10 cm = 0.1m
Let the Net electric field at the point O = E
The electric field at a point caused by charge q, is given as,
Where, q is the charge,
r is the distance between the charge and the point at which the field is being calculated
is a constant and its value is 9x109 N m2 C-2
ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 (F/m)
∴ The electric field at the point O caused by qA = 3 μC
The direction of EA will be along OB.
Also, the electric field at O caused by qB = –3 μC
The direction of EB will be along OB.
Note: Both the field are acting in the same direction along OB, this is because the charges are of opposite nature, ∴ the force will be attractive.
Hence, we can sum them up to find the resultant electric field.
Enet = EA + EB
⇒ Enet = 2EA
∴ Enet = 5.4 × 102 N/C(along OB)
(b) The diagram is:
q = 1.5 × 10–9 C
The force that is experienced by q when placed at O, F.
FO = q × ENet
Where E is the Electric field at the point O.
∴ F = 1.5 × 10–9 C x 5.4 × 102 N/C
⇒ F = 8.1 x 10-7 N
The negative test charge will be repelled by the force placed at B and attracted by the force placed at A. Therefore, this force will be in the direction of OA.
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There is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by Frr =
where the distance r is measured in cm (= 10–2 m), F in dynes (=10–5 N) and the charges in electrostatic units (es units), where 1es unit of charge 9 1 10 C  − = ×
The number  actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108 m/s. An approximate value of c then is c =  × 108 m/s.
(i) Show that the coloumb law in cgs units yields
1 esu of charge = 1 (dyne)1/2 cm.
Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.
(ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives
9 2 2 2 0 1 10 N.m 4x C π − = ∈
With , we have 9 1 x 10  − = ×
2 2 9 2 0 1 Nm  10 4C π = × ∈
Or, (exactly) 2 2 9 2 0 1 Nm [2.99792458] 10 4C π = × ∈
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(b) Assuming x << d, show that this force is proportional to x.
(c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement?
Find the time period of the oscillations if these conditions are satisfied.
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