2013-01-30, 10:12 AM
I know you've already got an answer for both things, but I'm going to stick my neck in anyway, but you didn't really need to do the whole quadratic thing.
If you know the initial velocity (2644779m/s), the final velocity (0m/s) and distance travelled(0.069m) for a constant acceleration (-2.535e13m/s^2), you could just use d=(u+v)*t/2 and come up with a much simpler equation where you less likely to make a mistake calculating it (This happens an awful lot, it's best to avoid the problem wherever possible
)
t =2*d/(u+v) = 2*0.069/(2644779)=5.22e-8 s
And the second bit is fairly straightforward too.
E field due to a charge Q is just Q/(4*Pi*permittivity of free space*r^2), and in this case, r =2m. Occasionally, they might make it harder, and put it at some point (x,y,z) from the origin, in which case you just use Pythagoras theorem to find the radius and away you go. Multiple charges add like vectors.
If you know the initial velocity (2644779m/s), the final velocity (0m/s) and distance travelled(0.069m) for a constant acceleration (-2.535e13m/s^2), you could just use d=(u+v)*t/2 and come up with a much simpler equation where you less likely to make a mistake calculating it (This happens an awful lot, it's best to avoid the problem wherever possible
)t =2*d/(u+v) = 2*0.069/(2644779)=5.22e-8 s
And the second bit is fairly straightforward too.
E field due to a charge Q is just Q/(4*Pi*permittivity of free space*r^2), and in this case, r =2m. Occasionally, they might make it harder, and put it at some point (x,y,z) from the origin, in which case you just use Pythagoras theorem to find the radius and away you go. Multiple charges add like vectors.