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.