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Nady [450]
3 years ago
14

A quarter circle of radius a is centered about the origin in the first quadrant and carries a uniform charge of −Q. Find the x-

and y-components of the net electric field at the origin.
Physics
1 answer:
loris [4]3 years ago
6 0

Answer:

E_x = \frac{2kQ}{\pi R^2}

E_y = \frac{2kQ}{\pi R^2}

Explanation:

Electric field due to small part of the circle is given as

dE = \frac{kdq}{R^2}

here we know that

dq = \frac{Q}{\frac{\pi}{2}R} Rd\theta

dq = \frac{2Q d\theta}{\pi}

Now we will have two components of electric field given as

E_x = \int dE cos\theta

E_x = \int \frac{kdq}{R^2} cos\theta

E_x = \int \frac{k (2Qd\theta) cos\theta}{\pi R^2}

E_x = \frac{2kQ}{\pi R^2} \int_0^{90} cos\theta d\theta

E_x = \frac{2kQ}{\pi R^2} (sin 90 - sin 0)

E_x = \frac{2kQ}{\pi R^2}

similarly in Y direction we have

E_y = \int dE sin\theta

E_y = \int \frac{kdq}{R^2} sin\theta

E_y = \int \frac{k (2Qd\theta) sin\theta}{\pi R^2}

E_y = \frac{2kQ}{\pi R^2} \int_0^{90} sin\theta d\theta

E_y = \frac{2kQ}{\pi R^2} (-cos 90 + cos 0)

E_y = \frac{2kQ}{\pi R^2}

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Answer:

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Explanation:

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