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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.

What is the change in the potential energy of charge +q during this process?
(A) kQq/R
(B) kQqR/r12
(C) kQqR/r22
(D) kQq((1/r2)-(1/r1))
(E) kQq((1/r1)-(1/r2))

Respuesta :

Answer:

Option (D)

Explanation:

The formula for the potential energy between the two charges is given by

[tex]U=\frac{KQq}{r}[/tex]

where, r is the distance between the two charges.

In first case the distance between the two charges is r1.

The potential energy is

[tex]U_{1}=\frac{KQq}{r_{1}}[/tex]

In first case the distance between the two charges is r2.

The potential energy is

[tex]U_{2}=\frac{KQq}{r_{2}}[/tex]

The change in potential energy is

[tex]\Delta U = U_{2}-U_{1}[/tex]

[tex]\Delta U=\frac{KQq}{r_{2}}-\frac{KQq}{r_{1}[/tex]

[tex]\Delta U=KQq \times \left ( \frac{1}{r_{2}} -\frac{1}{r_{2}} \right )[/tex]

ajeigbeibraheem

The change in the potential energy of the charge +q during this process is [tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

Option D is correct.

What is the potential energy between two charges?

In an electrical circuit, the electric potential energy is the entire potential energy of a unit charge will have if it is positioned at any point in space.

The electric potential generated by a charge at any point in space is directly proportional to its magnitude and also varies inversely proportional to the distance out from the point charge.

Mathematically, it can be expressed as:

[tex]\mathbf{U = \dfrac{KQq}{r}}[/tex]

here;

  • r = distance between the two charges

So,

  • The distance between the two charges for the first scenario is: = r₁

The potential energy for the first scenario can be expressed as:

[tex]\mathbf{U_1 = \dfrac{KQq}{r_1}}[/tex]

The potential energy for the second scenario can be expressed as:

[tex]\mathbf{U_2= \dfrac{KQq}{r_2}}[/tex]

Therefore, the change in the potential energy is:

[tex]\mathbf{\Delta U =U_2 -U1}[/tex]

[tex]\mathbf{\Delta U = \dfrac{KQq}{r_2}- \dfrac{KQq}{r_1}}[/tex]

[tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

Learn more about electric potential energy here:
https://brainly.com/question/14306881

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