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A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, VB − VA.

Respuesta :

Answer:

Vb - Va = -30V

Explanation:

Given data

charge Q = 50 μC

kinetic energy E =  1.5 mJ

particle moves = 25 cm

to find out

the electric potential difference

solution

we have find potential difference VB − VA tha is Vab

we have energy point charge i.e

E =  QV

put value

1.5 × [tex]10^{-3}[/tex] = 50 × [tex]10^{-6}[/tex]  × Vab

Vab = 1.5 × [tex]10^{-3}[/tex] / 50 × [tex]10^{-6}[/tex]

Vab = 30 V

we know that

Vab = - Vba

so Vba = -30 V

Vb - Va = -30V

samueladesida43

The electric potential difference (VB - VA) of the particle that moves in a region where the only force is an electric force is 30V.

How to calculate electric potential difference?

The electric potential difference can be calculated using the following expression:

According to this question, the following information are given:

  • Charge Q = 50 μC
  • Kinetic energy E = 1.5 mJ
  • particle distance = 25 cm

The electric potential differencee have energy point charge i.e

E = QV

1.5 × 10-³ = 50 × 10-⁶ × Vab

Vab = 1.5 × 10-³ ÷ 50 × 10-⁶

Vab = 1.5/50 × 10-³+⁶

Vab = 0.03 × 10³

Vab = 30

Since; Vab = - Vba

Vba = -30 V

Therefore, the electric potential difference (VB - VA) = 30V

Learn more about electric potential difference at: https://brainly.com/question/9383604

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