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Nostrana [21]
3 years ago
11

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

f the pair of charges is 5.4 x 10^-8J. When the second charge is moved to point b, the electric force on the charge does -1.9x10^-8J of work. What is the electric potential energy of the pair of charges when the second charge is at point b? Please explain the answer briefly
Physics
1 answer:
likoan [24]3 years ago
3 0

Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it  W_{a-b}=-1.9*10^{-8}J

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}

Now  substitute the given values

So

U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J

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The correct answer is c) 28 m/s.
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Explanation:

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