I know im forgetting something simple

[bemigh]

Once again, hello...
Alright, i know I am forgetting something simple...

A singly-charged positive ion of mass 2.50 ´ 10–26 kg, initially at rest, is accelerated through a potential difference of magnitude 2.50 ´ 102 V.

(a) Calculate the speed of the ion after passing through this potential difference.

(b) The ion then enters a region of uniform magnetic field of magnitude 0.500 T. The direction of the magnetic field is perpendicular to the ion’s velocity. Calculate the radius of the path of the ion in the field.

Now, i can solve this... however, I am stumped by something... How can i find the velocity of the particle after it leaves the potential difference... I would need q of the particle, allowing me to use K=qV, i know everything but q... is their another way of solving for the velocity??
Cheers

 

[StatusX]

No, you absolutely need q. How else could you calculate the electric force? since this is a singly charged ion, q is the basic charge e, 1.6x10^-19 C.

 

[thepatientmental]

!

Hello again! It seems like you are on the right track with solving the problem. To find the velocity of the ion after passing through the potential difference, you can use the equation for kinetic energy: KE = 1/2 mv^2. Since the ion is initially at rest, its initial kinetic energy is zero. Therefore, the final kinetic energy will be equal to the work done by the potential difference, which is qV. So, you can set up the equation as KE = qV and solve for v. Once you have the velocity, you can then move on to part (b) and solve for the radius of the path using the equation r = mv/qB. I hope this helps and good luck with solving the problem!