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Dark asked in Science & MathematicsPhysics · 2 months ago

Help on Physics?

So, I just am having trouble understanding why the particle cannot be placed at either position A or B on both parts of the question shown. I have provided an explanation on the picture by using the formula for forces between charges, but, it's not correct?

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  • NCS
    Lv 7
    2 months ago
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    Before I tackle your question, let's talk about the forces you drew at the top of the image. They're a little ambiguous.

    In terms of the forces that q and Q exert on each other, they are correct. But in terms of the context of the question, they shouldn't be there. The question is asking about the forces that q and Q exert on a THIRD particle.

    case 1: q negative and Q positive

    If that third particle is between q and Q and positive, then both forces point to the left.

    If that third particle is negative, then both forces point to the right.

    It is precisely because of this that when q and Q have opposite sign, there is no equilibrium position for a third charge BETWEEN q and Q.

    Only point A is possible.

    Note that E isn't possible for a different reason. If Q has greater magnitude than q, then the proximity of E to Q will have Q dominate the force regardless of the sign on the charge at E.

    case 2: q and Q both negative

    Reverse the logic above. Now the forces on the third particle will have opposite directions (inward as you have drawn them IF the third charge is negative).

    On the other hand, if the third charge is OUTSIDE of q and Q, then the forces point in the same direction and there can be no equilibrium. So the equilibrium point MUST BE between q and Q.

    But since Q has the larger magnitude, if the third charge is closer to Q than to q, then Q's force will always dominate. The third charge MUST BE closer to the charge of smaller magnitude so that its smaller distance can compensate.

    Summary: It is true that for q < 0 and Q > 0 there is a point B between them where the MAGNITUDES of the forces due to q and Q are the same. But they point in the same direction (and that direction depends on the third charge), and so there is no equilibrium at B.

    Similar argument for q and Q < 0 (or both > 0) -- outside the two charges there is a point where the MAGNITUDE due to each is the same, but again no equilibrium because they don't oppose each other.

    Conclusions: Q larger magnitude and opposite sign to q,only A is possible.For Q larger magnitude and same sign as q, only B is possible.

    Questions?

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