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Question:

A particle having a charge 10 mC is held fixed on a horizontal surface. A block of mass 80 g and having charge stays in equilibrium on the surface at a distance of 3 cm from the first charge. The coefficient of friction between the surface and the block is μ=0.5. Find the range within which the charge on the block may lie

8×10⁻⁵C to 4×10⁻⁵C

6C to 2×10²C

8×10⁻⁶C to 4×10⁻⁶C

6×10⁻⁶C to 2×10⁻⁶C

Solution:

Maximum force that can act due to friction is mgμ, which is equal to 0.4 N. Force due to charge has to be less than 0.4 N. Now force between the two is F=Kq₁q₂/r²=9×10⁹×0.01q/0.03²<0.4. Above equation implies charge has to be less than 4×10⁻⁵C, so option 1 is best.