Question

A proton is moving along the axis of a uniformly charged thin ring, of 10 nC/cm. The radius of the ring is 10 cm. If the proton started initially at 20 cm from the center of the ring with an initial velocity towards the ring, what initial speed does it need to be able to cross the plane of the ring?

Answer 1

charge in the ring,q = 10 * 10^-9 C * 2*pi* 0.1 = 6.28 * 10*^-9 C

F=q'E,

E = kq * x / (x^2 + r^2)^3/2...here x=20cm and r= 10 cm.. convert in metres

q' is charge of proton

acceleration of the proton,a = q'E/m .

now v^2 = u^2 + 2as

v=0 as the proton has to just reach

a is the acceleration just found above

and s=20cm or 0.2 m

Thus v can be found easily

U is the initial velocity which has to be calculated, V is the final velocity(=0)

Answer 2

Due to Sila changes, poton = 10nC. Ca Congeed 20 cm Y Just CYoming the plane at He Cenbre in uoill be Zeo et uhial sperd be U 6,283 X 1 0-1 C