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Anton [14]
2 years ago
8

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

find the magnitude and direction of the force this field exerts on a charge of +6.80 μc.
Physics
1 answer:
denis23 [38]2 years ago
5 0
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
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You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

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the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

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We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0
2 years ago
The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)
nikdorinn [45]

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E = 55.9583\ Volts/meter

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First let's find the electric potential using y = 22.5:

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E = 1259.0625/22.5

E = 55.9583\ Volts/meter

3 0
2 years ago
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