Question

In: Physics

Two charges Q1( +2.00 µC) and Q2( +2.00 µC) are fixed symmetrically along the x-axis at...

Two charges Q₁( +2.00 µC) and Q₂( +2.00 µC) are fixed symmetrically along the x-axis at x = ± 3.00 cm.

How much work must be done to assemble these charges if they are initially infinitely far apart?
For this configuration, calculate the electrical potential difference between the origin and the point P, which is located on the y-axis 4.00 cm above the origin.
Now a third charge Q₃( -2.00 µC) is fixed at point P. What is the electrostatic potential energy of this new configuration?
If Q₃ is released from rest, what is its speed when it reaches the origin if its mass is 0.100 g?

Expert Solution

a] Work done to introduce the 1st charge is 0 J since no forces are present.

Work done to introduce the 2nd charge = Work done to assemble both the charges = Potential Energy of the system

=>

b] Electric Potential at P =

Electric Potential at O =

so, Potential difference between O and P is: .

c] Electrostatic Potential Energy of the new configuration is:

where p = [3² + 4²] = 5 cm = 0.05 m

=>

d]

Change in Potential energy = change in kinetic energy

since the charge was released from rest,

=> v = 183.30 m/s

this is the speed of Q3 when it reaches the origin.

genius_generous answered 2 years ago

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