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A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a straight line to a new position at a distance R away from its starting position. The final location of q is at a distance r2 from Q. 1)What is the change in the potential energy of charge q during this process

Respuesta :

Answer:

[tex]\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})[/tex]

Explanation:

The electrostatic potential energy is given by the following formula

[tex]U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}[/tex]

Now, we will apply this formula to both cases:

[tex]U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}[/tex]

So, the change in the potential energy is

[tex]\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})[/tex]

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