A positive charge of 5.50 C is fixed in place. From a distance of 4.20 cm a part

Question

A positive charge of 5.50 C is fixed in place. From a distance of 4.20 cm a particle of mass 6.00 g and charge +3.10 C is fired with an initial speed of 72.0 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

q1=5.5 C
q2=3.1 C
d=4.2 cm=0.042 m
v=72m/s
m=6g=6*10^-3 kg

electric field by q1 is
E1=k*q1/d^2=2.80*10^7 N/C

let the q2 charge cover a distance of x m
work done by q2 to cover x m=work done by force produced by E1 to oppose the motion of
q2

0.5mv^2=F*x
mv^2=2*E1*q2*x
x=0.179 m= 17.9 cm

it will pass the q1 charge & it will be 17.9-4.2=13.7 cm near to q1

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