Recall that a cube has eight corners and six square A point charge Q, placed at the center of a cube, produces an electric field strength of

Question

Recall that a cube has eight corners and six square A point charge Q, placed at the center of a cube, produces an electric field strength of E — 130V/m, at each of the eight corners of the cube. What is the strength of the electric field at the centers of the six square faces of the cube?

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Thái Dương 2 months 2021-07-22T20:06:55+00:00 1 Answers 2 views 0

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    2021-07-22T20:08:09+00:00

    First we need to determine the distance from the center of the box to each of the corners. Geometrically this distance can be defined as

    d_1= \sqrt{ (\frac{a}{2})^2+(\frac{a}{2})^2+(\frac{a}{2})^2}

    d_1 = \frac{\sqrt{3}}{2}a

    Distance between center of cube and center of face is,

    d_2 = \frac{a}{2}

    Then the Electric field will be given for each point as,

    E_1 = \frac{kQ}{d_1^2}

    E_2 = \frac{kQ}{d_2^2}

    The relation between the two electric field then will be

    \frac{E_1}{E_2} = \frac{d_2^2}{d_1^2}

    The first electric field in function of the second electric field is

    E_2 = \frac{d_1^2E_1}{d_2^2}

    Replacing,

    E_2 = \frac{3}{4}*4 E_1

    E_2 = 3E_1

    Replacing with the value given,

    E_2 = 3*(130V/m)

    E_2 = 390V/m

    Therefore the strength of the electric field at the centers of the six square faces of the cube is 390V/m

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