Question

Two point charges of 1.0 microcoulomb and -0.25 microcoulomb are placed in air at a distance of 0.40 metre from each other find out at what point on the line joining the two charges should third charge be placed so that no force act upon it

Answer 1

Answer:

outsidethe second charge at a distance of.40m

Answer 2

Concept:

• We have to use the concept of electrostatic forces of attraction

Given:

• The first charge Q1 = 1 microcoulomb
• The second charge Q2 = -0.25 microcoulomb
• The distance between charges Q1 and Q2 = 0.4m

Find:

• The point on the line joining the charges where a thrid charge Q3 can be placed so that there is no net force acting on it

Solution:

• Assume that the distance between Q1 and Q3 is x
• Assume that the distance between Q2 and Q3 is x+0.4

F1 = force between charges Q1 and Q3

F1 = kQ1Q3/x^2

F2 = force between charges Q2 and Q3

F2 = kQ2Q3/(x+0.4)^2

For net force to be zero on Q3, F1=F2

kQ1Q3/x^2 = kQ2Q3/(x+0.4)^2

Q1/x^2 = Q2/(x+0.4)^2

1/x^2 = -0.25/(x+0.4)^2

-4 = x^2/(x+0.4)^2

2 = x/x+0.4

2x+0.8 = x

x = -0.8

Distance = 0.8+ 0.4 = 1.2

The third charge should be placed 1.2m away from the origin.

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