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NARA [144]
3 years ago
11

Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to

tal electric force that these charges exert on a third point charge Q = μC at x = m, y = 0?
Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

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Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

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Using formula of angular frequency

\omega=2\pi f

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We need to calculate the maximum value of the induced magnetic field that occurs at r = R

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B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

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V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

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