In case A two negative charges which are equal in magnitude are separated by a distance d. In case B the same charges are separated by a dis

Question

In case A two negative charges which are equal in magnitude are separated by a distance d. In case B the same charges are separated by a distance 2d. Which configuration has the highest potential energy

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Thu Giang 6 months 2021-07-29T09:26:48+00:00 1 Answers 6 views 0

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    2021-07-29T09:28:34+00:00

    Answer:

    Case A

    Explanation:

    We are given that

    Case A:

    Let

    q_1=q_2=-q

    Distance between two negative  charges,r=d

    Case B:

    Charges are same

    Distance between two charges,r=2d

    Potential energy=U=\frac{kq_1q_2}{r}

    Using the formula

    For case A

    U_A=\frac{kq^2}{d}

    For case B

    U_B=\frac{Kq^2}{2d}

    Potential energy is inversely proportional to distance between two charges

    The distance between two charges in case A is small therefore, it has the highest potential energy.

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