Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 years ago

Physics 2 Problem?

Two charges, -3Q and +Q, are fixed in place at points A and B with a separation ‘d’. Other than at infinite distance, how many points are there on the line through the charges at which the total electric potential is zero? Locate the points relative to the point A.

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  • 5 years ago
    Best Answer

    qa = -3Q

    qb = Q

    U = k*q/r

    For (2) points

    U = k*(qa/ra + qb/rb)

    And we want to find the 0

    0 = k*(qa/ra + qb/rb)

    0 = qa/ra + qb/rb

    There are (3) different locations you need to check. The problem looks something like this

    P1............ A ......... P2 ......... B ............... P3

    First, let's see if there is a point P1

    P1.......... A .......... B

    |---- x------|---- d ----|

    ra = x

    rb = x+d

    Out potential equation becomes:

    0 = -3Q/x + Q/(x+d)

    Q's cancel too

    0 = -3/x + 1/(x+d)

    3/x = 1/(x+d)

    3*(x+d) = x

    3x + 3d = x

    2x = -3d

    x = -3/2*d

    The negative means we guess the wrong direction. It should look like this

    A........... B........ P1

    |---- d ----|

    |-------- 3/2*d ------|

    So, the Potential can't be 0 on the LHS of A, which makes sense because magnitude of A is much larger than B and in the opposite direction.

    Let's check for the point inbetween

    A........ P2 ........ B

    |---- x ---|-- x-d --|

    |----------- d ------|

    0 = -3/x + 1/(x-d)

    3/x = 1/(x-d)

    3*(x-d) = x

    3x - 3d = x

    -3d = -2x

    x = 3/2*d

    Again, this puts us back in the same spot.

    So, it looks like the only point is 3/2*d away from A towards B.

    This makes sense. Say one was to put a point charge at any spot on the LHS of A. A is so much stronger than B, that the force from A will always have a net result on the charge. Since there would be a net force on the charge, it would automatically gain KE and that shouldn't happen when U = 0.

    Inbetween A and B, because of the opposite charges, they would both push/pull the charge in the same direction, so there would be a net force on the charge.

    On the RHS of B, which is where we found the point, even though B is weaker than A, when the charge gets closer to B it will experience a stronger reaction to B than to A. Since the charges are opposite, then A can push on the charge with the same force the B pulls on it (or vice versa depending on the +/- of the charge).

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