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tresset_1 [31]
2 years ago
10

2. Two charges 160μC and -250μC are placed together at a distance of 150 centimeter apart. Find the force of attraction between

each other. Also explain the nature of the force [Repulsing force = +ve, Attraction force = -ve]
​
Engineering
1 answer:
n200080 [17]2 years ago
4 0

The answer is -0.016 N, and as it negative, the nature of the force is attractive.

<u>Finding the force of attraction</u>

Coulomb's Law is given by :

\boxed {F = \frac{kq_{1}q_{2}}{r^{2}}}

<u>Solving</u>

  • F = 9 × 10⁹ × 160 × 10⁻⁶ × -250 × 10⁻⁶ / (150 × 10⁻³)²
  • F = -360000 × 10⁻³ / 22500 × 10⁻⁶
  • F = -16 × 10⁻³
  • F = -0.016 N
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In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

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   """

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   let learning rate = 1/sqrt(t),

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