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3 years ago

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we measure a voltage difference of 5.0 V between two points on the conducting paper between two parallel conducting electrodes.

If these two points are separated by 3.0 mm in a direction 24° with respect to the perpendicular to the electrodes, what is the magnitude of electric field at this location on the conducting paper and in which direction does the field point?

1 answer:

Alex17521 [72]3 years ago

8 0

Answer:

E=1824.81 V/m

Explanation:

Given that

Voltage difference = 5 V

Distance ,D= 3 mm

θ = 24°

As we know that electric filed given as

$E=\dfrac{V}{d}$

Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.

d= D cosθ

d= 3 cos24°

d = 2.74 mm

So

$E=\dfrac{V}{d}$

$E=\dfrac{5}{2.74\times 10^{-3}}$

E=1824.81 V/m

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Answer:

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Explanation:

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