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,


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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

(a) Describe briePhysics - Board Papers

Repeat the previoHC Verma - Concepts of Physics Part 2

There is another Physics - Exemplar

Does the charge gPhysics - Board Papers