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lisabon 2012 [21]
3 years ago
12

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge? * 48
Physics
1 answer:
Crank3 years ago
8 0

Answer:

Fnet = 12 N

Explanation:

Force on a point charge due to another point charge = kq1q2 / d^2

Force on +32uC = due to + 20uC + due to -60uC

where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2

Net Force =

= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2}  ]

F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]

=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N

Fnet = 12 N

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Two 1.0 g balls are connected by a 2.0-cm-long insulating rod of negligible mass. One ball has a charge of +10 nC, the other a 4
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Answer:

  α = 5 10⁻³ rad / s²

Explanation:

For this exercise we can use Newton's second law for rotational movement, where the force is electric

             τ = I α

Where the torque is

             τ = F x r = F r sin θ

Strength is

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The moment of inertia of a small ball, which we approximate to a point is

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We replace

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The number 2 is because the two forces create the same torque

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Let's calculate

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