I know this is a long question, that\'s why I have offered so many points. Pleas

Question

I know this is a long question, that's why I have offered so many points. Please help, I will rate

Problem 20.35

IP A charge of 19.0?C is held fixed at the origin.

A) If a -7.00?C charge with a mass of 3.70g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin? units are in m/s

B) Suppose the -7.00?C charge is released from rest at the point x = 12(0.925m) and y = 12(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

I know it's Greater than the speed found in part A, But why??? please explain

C) Find the speed of the charge for the situation described in part B. units are in m/s

Explanation / Answer

A)let the speed be v.

then initial potential energy=final potential energy+final kinetic energy

now,

step1:

initial potential energy=9*10^9*q1*q2/d

where d=sqrt(0.925^2+1.17^2)=1.49 m

energy=-0.80255 J

step 2:

when distance is d/2, final potential energy=-1.6055

so now 0.5*m*v^2-1.6055=-0.80255

v=20.828 m/s

b)as distance is smaller, the change in potential energy is greater.

so kinetic energy increaes==> speed is greater .

c)let new speed is v.

step 1:

initial potential energy=9*10^9*q1*q2/d

where d=sqrt((0.5*0.925)^2+(0.5*1.17)^2)=0.7457 m

energy=-1.6052 J

step 2:

when distance is d/2, final potential energy=-3.2104 J

so now 0.5*m*v^2-3.2104=-1.6052

v=29.456 m/s

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