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defon
4 years ago
12

Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc

onhecida Q. ficando sujeita a uma forca de atração de modulo 18 N. Sabendo que o meio e o vácuo (K = 9,0.10^9 Nm^2C^-2), determine: a) a intensidade do campo elétrico no ponto P; b) a carga fonte Q. O ponto P está a 30cm dessa carga.
Physics
1 answer:
expeople1 [14]4 years ago
7 0

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

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- The length of bar, L = 6.0 m

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The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

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                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

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