I really would appreciate help, I dont even kniw hwere to start with this... Pro
ID: 2037949 • Letter: I
Question
I really would appreciate help, I dont even kniw hwere to start with this...
Problem 6: Two positive charges are placed on the x-axis as shown in the dlagram below, They are both at rest and have a mass of 2.0 g A Which exerts a larger force: the force that charge two exerts on charge 1 or the force that charge 2 exerts on charge 17 Explain. (-9 m, 0 m) (+ 16 m, 0 m) How much electric potential energy does the system have? Is this energy positive or negative? B. C. How much work would be required to move the q 6nC charge to point O, while q, remains fixed? Does this correspond to an increase or decrease in electric potential energy? D. If charge 2 is allowed to move (while charge 1 still remains fixed in place), at what speed would charge qi be moving as it passes through its original position (-9cm)? E. How fast will charge 2 be moving by the time l gets very far away?Explanation / Answer
1
a)
The electric force between two charges is directly propotional to product of charge and inversly related to square of the distance between the charges. hence the two charges apply equal and opposite force on each other. so the magnitude of force will be same for both the charges
b)
r = distance between the charges = 16 - (- 9) = 25 m
electric potential energy is given as
U = k q1 q2/r
U = (9 x 109) (8 x 10-9) (6 x 10-9)/(25)
U = 1.728 x 10-8 J
c)
Ui = initial electric potential energy = (9 x 109) (8 x 10-9) (6 x 10-9)/(25) = 1.728 x 10-8 J
when q2 is moved to origin
Uf = initial electric potential energy = (9 x 109) (8 x 10-9) (6 x 10-9)/(16) = 2.7 x 10-8 J
work done is given as
W = Uf - Ui = (2.7 x 10-8) - (1.73 x 10-8) = 9.7 x 10-9 J
this correspond to increase in electric potential energy since the distance between the charges has decreased.
d)
kinetic energy gained = change in electric potential energy
(0.5) m v2 = 9.7 x 10-9
(0.5) (2 x 10-3) v2 = 9.7 x 10-9
v = 3.1 x 10-3 m/s
e)
kinetic energy = initial electric potential energy
(0.5) m v2 = 2.7 x 10-8
(0.5) (2 x 10-3) v2 = 2.7 x 10-8
v = 5.2 x 10-3 m/s