Question

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?

Answer 1

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

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;

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

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