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If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what will be the magnitude E of the electric field due to the particle at distance 1.00 meter from the particle?

Respuesta :

onyebuchinnaji

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

Electric field strength

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;

[tex]E = \frac{F}{q} \\\\E = \frac{kq}{r^2}[/tex]

where;

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

[tex]E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q[/tex]

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

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