The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5×104 N/C. Determine the force

Question

The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5×104 N/C. Determine the force (use dimensional analysis!) and resulting acceleration of an electron?

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Diễm Thu 2 months 2021-07-30T01:17:59+00:00 1 Answers 5 views 0

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    2021-07-30T01:19:04+00:00

    Answer:

    (a) F = -4.01 * 10^{-15} N

    (b) a = 4.40 * 10^{15} m/s^2

    Explanation:

    Parameter given:

    Electric field, E = 2.5 * 10^4 N/C

    (a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

    Mathematically:

    F = qE

    Electric charge, q, of an electron = - 1.602 * 10^{-19} C

    F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N

    (b) This electrostatic force causes the electron to accelerate with an equivalent force:

    F = -ma

    where m = mass of an electron

    a = acceleration of electron

    (Note: the force is negative cos the direction of the force is opposite the direction of the electron)

    Therefore:

    -ma =  -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}

    Mass, m, of an electron = 9.11 * 10^{-31} kg

    => a = \frac{-4.01 * 10^{-15}}{-9.11 * 10^{-31}}\\\\\\a = 4.40 * 10^{15} m/s^2

    The acceleration of the electron is 4.40 * 10^{15} m/s^2

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