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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?

Respuesta :

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

[tex]E=\dfrac{V}{d}[/tex]

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

[tex]E=\dfrac{V}{d}[/tex]

[tex]E=\dfrac{5}{2.74\times 10^{-3}}[/tex]

E=1824.81 V/m

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