Question

Two tiny spheres carrying charges of 1 microC and 3microC are placed 8 cm apart in air. What is the coulomb force on a unit positive test charge placed at the mid-point of the joining the two charges?

(1) 9/16 × 10^7 N
(2)9/64 × 10^5 N
(3)9/4 × 10^5 N
(4)9/8 × 10^7 N


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Answer 1

Answer:

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, q1 = 1.5 μC

Magnitude of charge located at B, q2 = 2.5 μC

Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V1 and E1 are the electric potential and electric field respectively at O

V1 = Potential due to charge at A + Potential due to charge at B

Where, ∈0 = Permittivity of free space

E1 = Electric field due to q2 - Electric field due to q1

Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4 × 105 V m - 1. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

V2 and E2 are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

V2= Electric potential due to A + Electric Potential due to B

Electric field due to q at Z,

Electric field due to q2 at Z,

The resultant field intensity at Z,

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V and electric field is 6.6 ×105 V m -1

Answer 2

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Yes, Above Answer is Right It is verified by me.

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