Question

You place a proton at point P and release it. The proton is constrained so that it can only move along the y axis. How fast will it be moving when it is 10 m away from the +3Q charge? (Show work and include unit)

Answer 1

Given

q1 = 1*10^-6 C, q2 = +3*1.6*10^-6 C, q3 = 1.6*10^-19 C

initially q1,q2 and q1,q3,q2,q3 are at a distance of r = 5 m later the charge q3 is at a distance 10m from q2 and 11.18 m from q1

19.

potential is v = kq/r

at P , v1 = kq1/r1 =kq1/(sqrt(R^2+R^2))

v1 = kq1/sqrt(2) *R

v1 = 9*10^9*1*10^-6/sqrt(2)*5 V

V1 = 1272.79221 V

at P by 3Qis , v2 = kq2/r2 = kq2/R

v2 = kq2/R

v2 = 9*10^9*3*10^-6/5 V

V2 = 5400 V

the potential is V = V1+V2 = 1272.79221 +5400 V

v = 6672.79221 V

20.

energy stored is U = kq1*q2/r

U = 9*10^9*3*10^-6*10^-6/5 J

U = 0.0054 J

the potential enery of the system of three charges is

U = u12+U23+U13

U = k(q1*q2/r12 + q3*q2/r23+q1*q3/r13)

U = 9*10^9 ((1*10^-6 *3*10^-6) /(5) + (1.6*10^-19 *3*10^-6)/5 + (1.6*10^-19 *1*10^-6)/(sqrt(2)*5 ) J

U = 0.0054 J = 5.4 m J

21. now the proton is placed at 10 m from q2 on y axis and released to move along y axis only

so the potential at this position is  

V = V1+V2

V = W/q3

V = k.e /q3

k.e = V*q3

0.5*m*v^2 = V*q3

v^2 = 2*V*q3/m

v = sqrt( 2*V*q3/m ) m/s ----------------->(A)

now first calculating the potential at q3 with r = 10 m and sqrt(5^2+10^2) = sqrt(125) m = 11.18 m from q1

V = k(q2/10 + q1/11.18)

V = 9*10^9((3*1*10^-6/10)+(1*10^-6/11.18)) v

V = 3505 V

substituting the value of V in (A)

v = sqrt( 2*V*q3/m ) = sqrt((2*3505*1.6*10^-19)/(1.6*10^-27)) m/s

v = 837257.43 m/s