(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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