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Viktor [21]
3 years ago
6

La magnitud del campo eléctrico E entre dos placas paralelas es 8kN/C. Las placas están separadas 0.03m. ?Cuánto trabajo realiza

el campo eléctrico E al mover una carga de -3 x10-6C desde la placa negativa hasta la placa positiva?
Physics
1 answer:
olchik [2.2K]3 years ago
7 0
<h2>Respuesta:</h2>

Si tenemos dos placas paralelas y un campo electrico de 8kN/C entre ellas y además conocemos la separación de estas placas que es igual a 0.03m. Entonces, cuando una carga se mueve de un punto a a un punto b el trabajo que es realizado  sobre la carga por el campo eléctrico viene dado por la siguiente ecuación:

W_{a \rightarrow b}=q_{0}E(y_{a}-y_{b})

Dado que la carga se mueve desde la placa negativa hasta la placa positiva, entonces:

y_{a}

W_{a \rightarrow b}=q_{0}E(-d) \\ \\ W_{a \rightarrow b}=-q_{0}Ed

Finalmente, sustituyendo:

W_{a \rightarrow b}=-(-3 \times 10^{-6})(8\times 10^{3})(0.03) \\ \\ \boxed{W_{a \rightarrow b}=7.2\times 10^{-4}J}

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Donde se hizo el “siluetazo”, <br>Me aydan​
brilliants [131]

Answer:

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ivanzaharov [21]

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s

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(

1

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Is=25MR2...........(1)

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