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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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Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

<u>Given:</u>

  • Speed of the car, v = 38.0 m/s.
  • Speed limit of the highway, \rm v_o=75.0\ mi/h.

<h2><u>Converting the speed limit from mi/h to m/s:</u></h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.

Therefore,

\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.

Clearly,

\rm v_o

which means, the car driver is exceeding the given speed limit.

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