If the charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q i

Question

If the charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is? <br /><br />Really need it, with a good explanation, will appreciate such answer. ​

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Bình An 4 years 2021-07-27T02:06:35+00:00 1 Answers 1 views 0

Answers ( )

  1. thongdat2
    0
    2021-07-27T02:07:38+00:00
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    Let’s just suppose this situation, so it’s like this;

    Q ————— q ————— Q

    Now, acc. to the question, the system is equilibrium which means that possibly both the Q should be exerting some pressure on the middle charge q, so let’s suppose Q be +ve and q be -ve

    The charge that we have kept at a certain distance something like this;

    +Q ————— -q ————— +Q

    <––– d ––– > <––– d ––– >

    Now, we +Q will be exerting some force on the another +Q that it should be acc. to columbs law like this;

    F = k × Q1Q2/r²

    F = k × QQ/(2d)²

    F1 = k × Q²/4d²

    Also, when this +Q will exert force on -q, it could be written as;

    F = k × Q1Q2/r²

    F2 = k × Qq/d²

    From the question, we know that the system is a equilibrium which means that these two force F1 and F2 will be equal to each other;

    = k × Q²/4d² = k × Qq/d²

    • Eliminate the common values

    = Q/4 = q

    • Wait wait! The answer had still not came! remember this q is actually -q?

    = Q/4 = -q

    = -Q/4 = q

    So, the charge on q is -Q/4.

    ——————————————-

    Have already answered this question, but still!

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )