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3 years ago

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The electric field, generated by a point charge, has strength of 180,000 N/C at a point 1.50 cm from the point charge. What is t

he charge, if the field points towards the charge

1 answer:

cestrela7 [59]3 years ago

7 0

Given Information:

Electric field = E = 180,000 N/C

Distance = r = 1.50 cm = 0.015 m

Required Information:

Charge = q = ?

Answer:

charge = q = 4.5x10⁻⁹ C

Explanation:

The electric field is given by

E = kq/r²

Where k is the coulomb constant k = 9x10⁹ Nm²/C²

q = Er²/k

q = 180,000*(0.015)²/9x10⁹

q = 4.5x10⁻⁹ C

Therefore, the charge is 4.5 nC

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Answer:

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Explanation:

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$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

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$r = \sqrt{0.141} = 0.377m$

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Answer:

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Explanation:

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Explanation:

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