The figure shows an arrangement of two -4.5 nC charges, each separated by 5.0 mm

Question

The figure shows an arrangement of two -4.5 nC charges, each separated by 5.0 mm from a proton. If the two negative charges are held fixed at their locations and the proton is given an initial velocity v as shown in the figure, what is the minimum initial speed v that the proton needs to totally escape from the negative charges?

neeative charges are beld fixod a heir koadions and the proton is given an initial velocity v as shown in the figure, what is the misimum initial speedthat t proton needs to totally esape from the segative chags? 14-10

Explanation / Answer

Initial potential energy of the system

= potential energy of two -4.5 nC charge with proton

= 2 x k x (-4.5 x 10-9) x 1.6 x 10-19 / 5 x 10-3

= 2 x 9 x 109 x (-4.5 x 10-9) x 1.6 x 10-19 / 5 x 10-3

= 25.92 x 10-16 j

Now if V is the speed of the proton, then kinetic energy of the proton = 1/2 x m x V2

In order to escape the kinetic energy should be more than or equal to potential energy of the system

1/2 x m x V2 =  25.92 x 10-16 j

so, 1/2 x 1.672 x 10-27 x V2 = 25.92 x 10-16

so, v= speed of the proton = 1.76 x 106 m/s

so answer is option a.

all the best in the course work

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