# Suppose the secon

Given:

Charges: q1=-1.0×10-6 C, q2=2.0×10-6 C

Let the third charge q be at a distance x cm from q2. 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 ϵ0 is the permittivity of free space

k is the electrostatic constant

q1 and q2 are the magnitude of charges

r is the distance of separation between the charges

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Force on q due to q1 is given by:

Force on q due to q2 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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