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In case A two negative charges which are equal in magnitude are separated by a distance d. In case B the same charges are separated by a distance 2d. Which configuration has the highest potential energy

Respuesta :

lublana

Answer:

Case A

Explanation:

We are given that

Case A:

Let

[tex]q_1=q_2=-q[/tex]

Distance between two negative  charges,r=d

Case B:

Charges are same

Distance between two charges,r=2d

Potential energy=[tex]U=\frac{kq_1q_2}{r}[/tex]

Using the formula

For case A

[tex]U_A=\frac{kq^2}{d}[/tex]

For case B

[tex]U_B=\frac{Kq^2}{2d}[/tex]

Potential energy is inversely proportional to distance between two charges

The distance between two charges in case A is small therefore, it has the highest potential energy.

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