Two point charges are separated by a distance d = 12 cm. One charge, q_2, is rel
ID: 1530004 • Letter: T
Question
Two point charges are separated by a distance d = 12 cm. One charge, q_2, is released from rest. If both charges are identical and have a charge of +15 mu C and, after traveling 5 cm, q_2 has a speed of 6 m/s what is the mass of q_2? Now assume the charges have different charges, where q_1 = +6 mu C and q_2 = -9 mu C. When q_2 is released from rest and halfway to q_1 what will its speed be (using your mass from part A)? An isolated metal sphere has a charge density of -12 nC/m^3 and a radius of 10 cm. How much work would it take to bring q2 from part B from far away to the surface of the sphere? What would q_1 's (from part B) speed be if it was released from rest far away from the isolated sphere in part C when it reached the sphere's surface (Assuming it had the same mass as q_2)?Explanation / Answer
a) Initial Electroststatic PE = KE + Final Electroststatic PE
K ( 15 x 10^-6)( 15x 10^-6)/ ( 0.12) = K ( 15 x 10^-6)( 15x 10^-6)/ ( 0.07 ) + 0.5 ( m) ( 6)^2
2025 x 10^ -3 ( 8.33- 5.88 ) = 18 m
m( of q2) = 1.5 kg apprx
b) K ( 6 x 10^-6) ( -9 x 10^-6 )/ (0.12) = K ( 6 x 10^-6) ( -9 x 10^-6 )/ (0.06) + 0.5 ( 1.5) v^2
( -486/ 0.12 + 486/ 0.06) = 0.75 v^2
486x 10^-3 ( 8.33 )= 0.75 v^2
v = 2. 323 m/s apprx
c) let's find the charge on sphere= Q= charge density x volume = -12 x 10^-9 ( 4/3 pi x 0.01) = -0.118 x 10^ -9 C
Work done= 9x 10^ 9 ( -.0.118 x 10^-9 ) ( -9 x 10^-9 ) / ( 0.01) = 9.55 x 10^-7 J
d) We need to conserve the energy,
initial PE = 0 ( as the system of charges are seperated by infinity)
Final PE + KE = 0
9 x 10^ 9 ( -0.118 x 10^ -9) (6 x 10^-6 ) / ( 0.01 ) + 0.5 ( 1.5 ) v^2 = 0
637. 2 x 10^ -3 = 0.75 v^2
v = 0.9217 m/s apprx
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.