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Gnesinka [82]
3 years ago
15

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?
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

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

Explanation:

When a particle moves from a point where the potential energy is Ua to point where it is Ub,the change in potential energy is is equal to work done

So

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

Where Wa-b here is negative this means Ub is greater Ua. Therefore potential energy increases

So

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

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I have two (2) brilliant ideas:

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2). Somewhere inside the car, like under the dash or in the glove box, install a teeny tiny radio receiver that can recognize the signals coming OUT of your phone.  Connect it to the car's electrical system so that when it hears signals from phones inside the car, it it shuts down the car's motor so you can't start or drive. The car only works when phones inside the car are either turned off or in Airplane Mode.

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Answer:

See answer

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d\vec{A} is parallel to \vec{B} and \vec{B} is constant in magnitude and direction therefore:

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Part A)

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after the interval \Delta t= 2.4 [m/s]

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now, the EMF is defined as:

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if we consider \Delta t= 2.4 [m/s] very small then we can re-write it as:

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