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Myr73
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A charge of -4.00uc is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 X 10^-3kg and charge -3.00uc is fired with an initial speed of 15.0m/s directly toward the fixed charge. How far does the particle travel before it stops and begins to return back?
q1= -4X 10^-6C d= 0.55m m=2.5 X 10^-3 kg q2= -3 x 10^-6 C
V0= 15 m/s x=? x= 0.55-r {r being the distance in between charge 2's maximum distance traveled(x) and charge 1}
KE=PE Ke= 0.5mv^2 and then PE either is =q2V or = kQ1Q2/r
So I would either do a) 0.5mv^2=q1V --> V= 0.5(2.5X 10^-3)(15^2)/(-4 x 10^-6)
and then --> r= V/kQ --> x= 0.55-r
or I'm thinking more likely b) 0.5mv^2=kQ1Q2/r --> r=kQ1Q2/0.5mv^2
--> x=0.55-r
I think it would be b) but I am unsure and would I just use the mass given of 2.5 X 10^-3kg? the charges wouldn't have a mass I guess -
q1= -4X 10^-6C d= 0.55m m=2.5 X 10^-3 kg q2= -3 x 10^-6 C
V0= 15 m/s x=? x= 0.55-r {r being the distance in between charge 2's maximum distance traveled(x) and charge 1}
Homework Equations
KE=PE Ke= 0.5mv^2 and then PE either is =q2V or = kQ1Q2/r
The Attempt at a Solution
So I would either do a) 0.5mv^2=q1V --> V= 0.5(2.5X 10^-3)(15^2)/(-4 x 10^-6)
and then --> r= V/kQ --> x= 0.55-r
or I'm thinking more likely b) 0.5mv^2=kQ1Q2/r --> r=kQ1Q2/0.5mv^2
--> x=0.55-r
I think it would be b) but I am unsure and would I just use the mass given of 2.5 X 10^-3kg? the charges wouldn't have a mass I guess -