HELP An electron is launched horizontally to the right with an initial speed of 100 m/s in a uniform horizontal electric field of 10,000 v/m where field lines point from left to right. How long after launch until the electron's speed is zero?

Question

chupacabraj · Answer

f=ma, so use F=Eq

chupacabraj · Answer

for f  get 1.6 x 10^-15 N

chupacabraj · Answer

Using kinematics to solve for t I get a negative, and the absolute value of that isn't even the solution

irishboy123 · Answer

$a = \dfrac{10 \cdot 10^3 \times -1.6 \cdot 10^{-19}}{9.1 \cdot 10^{-31}} = -1.75 \cdot 10^{15}$

for the electron, $KE_o = 0.5 \cdot 10^{-26}$ and the energy radiated away during the deceleration is $\approx 10^{-46} $J, so that's insignificant.   IOW (non-relativistic) kinematics should give the answer

chupacabraj · Answer

Ohh I forgot about the negative sign since it's an electron!

chupacabraj · Answer

Vf = Vi + at?

chupacabraj · Answer

Is that the correct kinematic equation to use in order to get the time at which v= 0?

chupacabraj · Answer

This is the answer

irishboy123 · Answer

yes

$v_f = u_i + at$

$t  =  \dfrac{0-100}{−1.75⋅10^{15}} \approx 5.7 \cdot 10^{-14}$

check the physical situation first to avoid nasty surprises.  if the E field is left to right, a proton (+ve charge) will move left to right in that field.  that's how the E field vector is defined.  and so an electron will want to move right to left

getting the signs in the formula right is then easier

the answer book looks wrong :(

chupacabraj · Answer

It looks similiar, perhaps they decided to keep it in fraction form? I just wonder how they'd do it though. They won't let me use a calculator. :S