Coderhk
- 59
- 2
A charge of +2.5 micro coulomb is at the origin and a +3.5 micro coulomb is at the point (3,0). What is the velocity of a proton when it is at (6,0) if it was released at (5,0).
My solution:
$$E_0=E_f$$
$$PE=KE$$
$$Since...Work = -PE$$
I can calculate the work it takes to move the proton from (6,0) to (5,0) multiply that by negative one and set that equal to the kinetic energy.
To calculate the net work and I can calculate the work required to move the proton against each charge and add them together.
$$W=\sum\int{dw}=(\int_6^5{\frac{kq_1}{r^2}dr})+\int_6^5{\frac{kq_2}{r^2}dr}=-k(\frac{q_1+q_2}{30})$$
$$k(\frac{q_1+q_2}{30})=\frac{1}{2}mv^2$$
$$v=\sqrt{k(\frac{q_1+q_2}{15m})}$$
$$v=1.46E15$$
However my answer seems to be incorrect.How should I approach this question instead?
Correct answer is 1.1E6
My solution:
$$E_0=E_f$$
$$PE=KE$$
$$Since...Work = -PE$$
I can calculate the work it takes to move the proton from (6,0) to (5,0) multiply that by negative one and set that equal to the kinetic energy.
To calculate the net work and I can calculate the work required to move the proton against each charge and add them together.
$$W=\sum\int{dw}=(\int_6^5{\frac{kq_1}{r^2}dr})+\int_6^5{\frac{kq_2}{r^2}dr}=-k(\frac{q_1+q_2}{30})$$
$$k(\frac{q_1+q_2}{30})=\frac{1}{2}mv^2$$
$$v=\sqrt{k(\frac{q_1+q_2}{15m})}$$
$$v=1.46E15$$
However my answer seems to be incorrect.How should I approach this question instead?
Correct answer is 1.1E6