Electron Approaching a Charged Sphere

AI Thread Summary
An electron starting from rest at a distance of 2.80 cm from a uniformly charged sphere with a charge of 1.01×10^-9 C is analyzed for its speed upon reaching the sphere's surface. The relevant equations include the conservation of energy, where the initial potential energy and kinetic energy equal the final potential and kinetic energy. The calculations involve constants such as Coulomb's constant and the electron's charge and mass, leading to a derived speed of approximately 7.9×10^7 m/s. A question arises regarding the role of the term (r-initial - radius) in the calculations. Clarification on this factor is sought to resolve the solution process.
jrk012
Messages
7
Reaction score
0

Homework Statement



An electron starts from rest 2.80 cm from the center of a uniformly charged sphere of radius 1.83 cm. If the sphere carries a total charge of 1.01×10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?


Homework Equations



UI + KI = UF + KF
U=(k*Q*-e)/r
K=1/2mv^2


The Attempt at a Solution



Rearranging terms I got:

[([(k*Q*-e)/(r-initial)] - [(k*Q*-e)/(radius)])/(r-initial - radius)](2/m)

Using:
k = 9x10^9
Q = 1.01×10-9
e = 1.602 x 10^(-19)
m = 9.1094 x 10^(-31)
r-initial = .028m
radius = .0183m

I then get v^2 = 6.24x10^15. I square root(√) it to get v and I get:

v = 7.9x10^7 m/s

I have tried the negative and positive of this and I cannot solve it. Any help would be appreciated
 
Physics news on Phys.org
jrk012 said:

Homework Statement



An electron starts from rest 2.80 cm from the center of a uniformly charged sphere of radius 1.83 cm. If the sphere carries a total charge of 1.01×10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?


Homework Equations



UI + KI = UF + KF
U=(k*Q*-e)/r
K=1/2mv^2


The Attempt at a Solution



Rearranging terms I got:

[([(k*Q*-e)/(r-initial)] - [(k*Q*-e)/(radius)])/(r-initial - radius)](2/m)

Using:
k = 9x10^9
Q = 1.01×10-9
e = 1.602 x 10^(-19)
m = 9.1094 x 10^(-31)
r-initial = .028m
radius = .0183m

I then get v^2 = 6.24x10^15. I square root(√) it to get v and I get:

v = 7.9x10^7 m/s

I have tried the negative and positive of this and I cannot solve it. Any help would be appreciated
What is the factor, (r-initial - radius) doing in there?
 
SammyS, you are a saint. Thank you!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top