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13 January, 16:57

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at x 5 13a. The net electric field these charges produce at the origin has a magnitude of 2keQ / a2. Explain how many values are possible for the unknown charge and find the possible values.

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  1. 13 January, 16:58
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    Q_2 = + / - 295.75*Q

    Explanation:

    Given:

    - The charge of the first particle Q_1 = + Q

    - The second charge = Q_2

    - The position of first charge x_1 = 2a

    - The position of the second charge x_2 = 13a

    - The net Electric Field produced at origin is E_net = 2kQ / a^2

    Find:

    Explain how many values are possible for the unknown charge and find the possible values.

    Solution:

    - The Electric Field due to a charge is given by:

    E = k*Q / r^2

    Where, k: Coulomb's Constant

    Q: The charge of particle

    r: The distance from source

    - The Electric Field due to charge 1:

    E_1 = k*Q_1 / r^2

    E_1 = k*Q / (2*a) ^2

    E_1 = k*Q / 4*a^2

    - The Electric Field due to charge 2:

    E_2 = k*Q_2 / r^2

    E_2 = k*Q_2 / (13*a) ^2

    E_2 = + / - k*Q_2 / 169*a^2

    - The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

    E_net = E_1 + E_2

    2kQ / a^2 = k*Q_1 / 4*a^2 + / - k*Q_2 / 169*a^2

    - The two equations are as follows:

    1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

    2Q = Q / 4 + Q_2 / 169

    Q_2 = 295.75*Q

    2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

    2Q = Q / 4 - Q_2 / 169

    Q_2 = - 295.75*Q

    - The two possible values corresponds to positive and negative charge Q_2.
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