A positive charge of 4.60 µC is fixed in place. From a distance of 4.60 cm a par

Question

A positive charge of 4.60 µC is fixed in place. From a distance of 4.60 cm a particle of mass 5.70 g and charge +3.70 µC is fired with an initial speed of 68.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away?

(The particle comes to rest when it has no kinetic energy. The change in kinetic energy will be equal to the increase in electrical potential energy. Note that the particle has some potential energy at the starting point. )

Explanation / Answer

Initial energy of particle = .5 x 5.7x10-3 x68x68 + 4.6x3.7 X10-12 x9x109 /(4.6x10-2 )

Energy of the particle when it comes to rest= /d

Conservation of energy both should be equal

(4.6x3.7 X10-12 x9x109    )(1/d- 100/4.6)= .5 x 5.7x10-3 x68x68

(1/d- 100/4.6)= .5 x 5.7x10-3 x68x68/( 4.6x3.7 X10-12 x9x109 )

        

1/d-21.73=86   1/d=86+21.73 m d=.009 m=.9 cm

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