Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0

Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

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Dulcie 6 months 2021-08-09T06:57:21+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-09T06:59:20+00:00

    Complete Question

    Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

    Required:

    What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

    Answer:

    The velocity is v = 80.82 \ m/s

    Explanation:

    From  the question we are told that

       The magnitude of charge one is  q_1 =  3.25 nC  =  3.25 *10^{-9} \  C

       The  magnitude of charge two  q_2  =  2.00 \ nC  =  2.00 *10^{-9} \ C

       The distance of separation is   d =  58.0 \ cm  =  0.58 \ m

     

    Generally the electric potential of the electron at the midway point is mathematically represented as

             V  =  \frac{ q_1 }{\frac{d}{2} }  + \frac{ q_2}{\frac{d}{2} }

    substituting values

             V  =  \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} }  +  \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }

              V  =  1.8103 *10^{-8} \  V

    Now when the electron is 10 cm   =  0.10 m  from charge 1 , it is  (0.58 – 0.10 =  0.48 m ) m from charge two

    Now the electric potential  at that point is mathematically represented as

           V_1  =  \frac{q_1}{ 0.10}  +  \frac{q_2}{ 0.48}

     substituting values

          V_1  =  \frac{3.25 *10^{-9}}{ 0.10}  +  \frac{2.0*10^{-9}}{ 0.48}

          V_1  = 3.67*10^{-8}  \ V

    Now the law of energy conservation ,

       The  kinetic energy of the electron  =  potential energy of the electron

    i.e     \frac{1}{2} * m * v^2  =  [V_1 -  V]* q

    where q is the magnitude of the charge on the electron with value

          q = 1.60  *10^{-19} \ C

    While m is the mass of the electron with value  m = 9.11*10^{-31} \  kg

            \frac{1}{2} * 9.11 *10^{-19} * v^2  =  [ (3.67 -  1.8103) *10^{-8}]* 1.60 *10^{-19}

             v = \sqrt{6532.4}

            v = 80.82 \ m/s

     

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