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Alla [95]
2 years ago
5

If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what wi

ll be the magnitude E of the electric field due to the particle at distance 1.00 meter from the particle?
Physics
1 answer:
kicyunya [14]2 years ago
7 0

The magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

<h3>Electric field strength</h3>

The electric field strength of a charged particle is the force per unit charge in the given field.

The electric field strength of a charge is given as;

E = \frac{F}{q} \\\\E = \frac{kq}{r^2}

where;

  • k is Coulomb's constant
  • q is the charge
  • r is the distance

E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q

Thus, the magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

Learn more about electric field here: brainly.com/question/14372859

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