Q. 74.5( 11 Votes )
Suppose the secon
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.
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
Force on q due to q1 is given by:
Force on q due to q2 is given by:
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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