Answer :

(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,


…(1)


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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