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Masja [62]
3 years ago
11

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
Physics
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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For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

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                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

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       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

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