Question

A point charge of 6* 10 to the power of -9 C is placed at a distance of 30cm from another point-charge of 4*10 to the power of -9C.find out the intensity of electric field at a point between the charges on the line joining the charges and at a distance of 20 cm from the bigger charge

Answer 1

Answer:

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Explanation:

Answer 2

Concept:

The magnitude of the electric field is expressed with the formula E = Kq/r², where E is the strength of the electric field, K is the coulombs constant , and q is the charge, r is the distance.

Given:

Point charge 1 = 6 × 10⁻⁹ C

Point charge 2 = 4 × 10⁻⁹ C

Distance between them = 30 cm or 0.3 metres

Find:

Electric field in between both charges with 20 cm or 0.2 metres from bigger magnitude charge.

Solution:

Net Electric field = Electric field by charge 1 at 0.2 metres - Electric field by charge 2 at 0.1 metres

we know value of k is 9×10⁹

Electric field by charge 1 at 0.2 metres = Kq/r²

Electric field by charge 1 at 0.2 metres = (9×10⁹) × (6 × 10⁻⁹) / 0.2²

Electric field by charge 1 at 0.2 metres = 54 / 0.04

Electric field by charge 1 at 0.2 metres = 1350 V/m

Electric field by charge 2 at 0.1 metres = Kq/r²

Electric field by charge 2 at 0.1 metres = (9×10⁹) × (4 × 10⁻⁹) / 0.1²

Electric field by charge 2 at 0.1 metres = 36 / 0.01

Electric field by charge 2 at 0.1 metres = 3600 V/m

Net Electric field = Electric field by charge 2 at 0.1 metres - Electric field by charge 1 at 0.2 metres

Net Electric field = 3600 - 1350 V/m

Net Electric field = 2250 V/m

Hence the magnitude of net electric field is 2250 V/m.

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