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11 December, 11:26

An electron is acted on by two electric forces, one of 2.7*10-14 N acting upward and a second of 5.8*10-14 N acting to the right. What is the magnitude of the electric field at the electron's location?

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  1. 11 December, 11:38
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    3.99*10⁵ N/C

    Explanation:

    Assuming both force forms a right angle triangle,

    Using Pythagoras theorem,

    a² = b²+c² ... Equation 1

    Where a = resultant of the two forces, b = the force acting upward, c = the force acting to the right.

    make a the subject of the equation.

    a = √ (b²+c²) ... Equation 2

    Given: b = 2.7*10⁻¹⁴ N, c = 5.8*10⁻¹⁴ N.

    Substitute into equation 2

    a = √[ (2.7*10⁻¹⁴) ² + (5.8*10⁻¹⁴) ²]

    a = √[7.28*10⁻²⁸) + (33.64*10⁻²⁸) ]

    a = √ (40.92*10⁻²⁸)

    a = 6.397*10⁻¹⁴ N.

    But,

    F = Eq ... Equation 3

    Where E = Electric Field, q = charge of an electron, F = electric force.

    make E the subject of the equation,

    E = F/q ... Equation 4

    Given: F = 6.397*10⁻¹⁴ N, q = 1.602*10⁻¹⁹ C.

    Substitute into equation 4

    E = 6.397*10⁻¹⁴ / (1.602*10⁻¹⁹)

    E = 3.99*10⁵ N/C
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