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Alex17521 [72]
2 years ago
15

Two identical balls balls P and Q, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal length

. The balls position themselves at equilibrium such that the angle between the threads is 60_{o}.If the distance between the the balls is 0.5m, find the charge on each of them. Take g = 9.8 m/s^{2}

Physics
1 answer:
Elden [556K]2 years ago
3 0
The charge on each of them is 0.173μC

Given the angle between the two  threads is 60° and the length of the two threads are equal.So, it forms an isoceles triangle so the other two angles will also be equal to 60°. Now,

Now, take a ball and we see that three forces act on it:

(i) Tension = T

(ii) Weight = mg

(iii) Electric force =  F (this will be replusive force because charges are same)

Tension T can be written as :

For equilibrium conditons, F = Tcos60°, mg = Tsin60°

Dividing, we get, \frac{F}{mg} = cot 60°

And we know, electric force , F = k\frac  {q^{2}}{r^{2}}

thus, q = \sqrt{ \frac{cot 60 . 9.8 . 0.20 * 10^{-3} * 0.25 }{9 * 10^{9} } 

          =  1.73 ×10^{-7}

         =  0.173μC

                                               

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A) Calculate the resultant electric field strength at the midpoint between the charges.

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E at midpoint = k×Qx/0.25² - k×Qy/0.25²

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Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

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C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

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∠Y = tan⁻¹(0.4/0.3) = 53.13°

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Determine the horizontal component of E at P:

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Determine the vertical component of E at P:

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Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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