Two charges at the ends of a rod .

The figure shows a long, nonconducting, massless rod of length L L , pivoted at its center and balanced with a block of weight W W at a distance x x from the left end. At the left and right ends of the rod are attached small conducting spheres with positive charges q q and 2 q 2q , respectively. A distance h directly beneath each of these spheres is a fixed sphere with positive charge Q Q . The value h h should have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced can be expressed as a q Q 4 π ϵ 0 W \sqrt{ \frac{a qQ}{4 \pi \epsilon_{0} W}} what is the value of a + 1 a+1 .


The answer is 4.

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2 solutions

Sravan Kumar
Nov 18, 2014

all vertical forces are balanced so that the repulsive force(total) is equal to the weight of the block W=(1/4pi epsilon){3Qq/h^2}

since the rod is long, we can ignore the forces between diagonal pairs of charges. since the bearing exerts no vertical force on the rod, we can say that the weight W is equal to the sum of the forces of electrostatic repulsion. we get: 3 qQ / (4 pi epsilon h^2) = W. solving for h, we get that a + 1 = 4.

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