The figure below shows a long, nonconducting, massless rod oflength L , pivoted

Question

The figure below shows a long, nonconducting, massless rod oflength L, pivoted at its center and balanced with a blockof weight W at a distance x from the left end. Atthe left and right ends of the rod are attached small conductingspheres with positive charges q and 2q,respectively. A distance h directly beneath each of thesespheres is a fixed sphere with positive charge Q.(a) Find equation for the distance x whenthe rod is horizontal and balanced. (b) What valueshould h have so that the rod exerts no vertical force onthe bearing when the rod is horizontal and balanced? Usepermittivity constant 0.

Explanation / Answer

Yeah... this is a silly question. But I've seen much worse. Bythat, I mean I've seen really really stupid physics questions thatmake you want to just smack the person who wrote them and tell them"GET A LIFE!" . Anyway... this one isn't so bad. It involves just twothings:   force between two point charges and torques in equilibrium. . There are three torques acting on the rod:   .             from the force of the charges at the left end (repulsiveforce, tries to turn the rod clockwise) .             from the weight (tries to turn the rod clockwise) .            from the force of the charges at the right end (repulsive force,tries to turn the rod counterclockwise) . So...  each torque is just the force times thedistance from the center of the rod. Therefore: .     clockwisetorques:      (1/2) L * k Q q/ h2      and      W (x - L/2) .     counterclockwisetorque:        (1/2) L * kQ 2q / h2 . They are balanced, so     (1/2) L * k Q q / h2     +    W (x - L/2)   =       (1/2) L * k Q 2q /h2 .    solve forx:           W (x - L/2) = (1/2) L * k Q q /h2      .             x- L/2 =      L k Q q / 2 Wh2 .          x =    (L/2) ( 1 + k Q q / Wh2 )       also,   k is 1/4o    so .          x = (L/2) ( 1 + Q q /4o W h2 ) . For part (b)... they want you to balance the threeforces, so: .      up forces = downforces .       k Q q /h2     +     k Q 2q /h2      =     W .        3  k Q q /h2        = W .          h2 =   3 k Q q / W =    3 Q q / 4oW      or .         h = (  3 Q q / 4oW   )1/2 .          x =    (L/2) ( 1 + k Q q / Wh2 )       also,   k is 1/4o    so .          x = (L/2) ( 1 + Q q /4o W h2 ) . For part (b)... they want you to balance the threeforces, so: .      up forces = downforces .       k Q q /h2     +     k Q 2q /h2      =     W .        3  k Q q /h2        = W .          h2 =   3 k Q q / W =    3 Q q / 4oW      or .         h = (  3 Q q / 4oW   )1/2

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