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27 October, 18:34

A positive charge Q1 = 34 nC is located at the origin. A negative charge Q2 = - 5.5 nC is located on the positive x-axis p = 15 cm from the origin. Calculate the location of x axis in cm where electric field is.

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  1. 27 October, 19:36
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    10.7 cm and 25.1 cm

    Explanation:

    Given:

    Q_1 = 34 * 10^ (-9) C

    Q_1 = - 5.5 * 10^ (-9) C

    Distance between charges = 0.15 m

    Distance from origin where E is zero = x

    k = 8.99*10^9 Nm^2 / C^2

    It is to be noted that E due to positive charge at origin is directed along + x-axis while for negative charge it is directed towards the origin. General, the electric field strength E at a distance R due to charge q is given by:

    E = k * (Q) / R^2

    The Net Electric field strength of the two charges and is equated to zero as follows:

    E_net = k * (Q_1) / x^2 + k * (Q_2) / (0.15 - x) ^2 = 0

    (Q_1) / x^2 + (Q_2) / (0.15 - x) ^2 = 0

    Solve the above equation for x:

    Q_1 * (0.15 - x) ^2 + Q_2 * x^2 = 0

    Q_1 * (x^2 - 0.3x + 0.0225) - Q_2 * x^2 = 0

    (Q_1 + Q_2) x^2 - 0.3*Q_1*x + 0.0225*Q_1 = 0

    2.85x^2 - 1.02x + 0.0765 = 0

    x = 0.107 m

    x = 0.251 m
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