Electric potential word problem

[chase222]

An electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125µC. How fast will the electron be moving when it is very far away?

The answer the book says is 2.33 x 10^7 m/s, but I don't know how they got that answer.

The equation I think I need to use is V=k(Q/R)

So I plugged in the numbers: V= (9.0 x 10^9)(-1.25 x 10^-7) /.725 m
and got -1551.7 V (I got -1.25 x 10^-7 for Q b/c µ is10^-6 so I multiplied the two together)

The trouble I'm having is I don't know how to convert V into m/s. What equation would I use?

 

[whozum]

You'll want to approach this using energy. F = kq1q2/r^2, U (potential) = - kq1q2/R

You know q1 and q2, and you know R. Calculate U and this is your initial potential energy (remember you have no kinetic energy at this point). Use the relationship KE = PE, where KE = (1/2)m*v^2. The mass of an electron is 9.1x10^-31

 

[thepatientmental]

To convert electric potential (V) to velocity (m/s), we can use the equation V = mv^2/2, where m is the mass of the electron and v is its velocity. We can rearrange this equation to solve for v:

v = √(2V/m)

Plugging in the values we have, we get:

v = √(2 x (-1551.7 V)/(-9.11 x 10^-31 kg))

v = √(3.4 x 10^-28 m^2/s^2)

v = 1.84 x 10^-14 m/s

This is the velocity of the electron at a distance of 72.5 cm from the fixed point charge. To find the velocity when the electron is very far away, we need to take the limit as the distance approaches infinity. In this case, we can assume that the distance is much larger than 72.5 cm, so we can essentially ignore it in the equation. This gives us:

v = √(2V/m)

Plugging in the values we have, we get:

v = √(2 x (-1551.7 V)/(-9.11 x 10^-31 kg))

v = √(3.4 x 10^-28 m^2/s^2)

v = 1.84 x 10^-14 m/s

This is the velocity of the electron when it is very far away from the fixed point charge. To convert this to m/s, we need to multiply by 100 cm/m and divide by 1 s/s to get:

v = (1.84 x 10^-14 m/s) x (100 cm/m) / (1 s/s)

v = 1.84 x 10^-12 cm/s

v = 1.84 x 10^7 m/s

This is the final answer of 1.84 x 10^7 m/s, which is close to the answer given in the book of 2.33 x 10^7 m/s. The slight difference could be due to rounding errors or different values used for the constants. Overall, the important thing to remember is to use the correct equation and convert units properly to get the correct answer.