Q. 74.5( 11 Votes )

# Suppose the secon

Answer :

__Given:__

Charges: q_{1}=-1.0×10^{-6} C, q_{2}=2.0×10^{-6} C

Let the third charge q be at a distance x cm from q_{2}. As the forces must cancel, case II is not possible. Moreover, for forces to cancel, q must be near the smaller charge. Hence, only Case III is possible.

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**Formula used:**

By Coulomb’s law, the electric force is given by:

Where ϵ_{0} is the permittivity of free space

k is the electrostatic constant

q_{1} and q_{2} are the magnitude of charges

r is the distance of separation between the charges

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Force on q due to q_{1} is given by:

Force on q due to q_{2} is given by:

Now,

For Case III, x must be greater than 10. Hence, x = 34.14 cm

Thus, q should be placed *34.1 cm* from the larger charge on the side of the smaller charge.

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