Respuesta :

onyebuchinnaji

Answer:

The strength of the electric field is 29922.34 N/C

Explanation:

Given;

distance of the electric field, d = 5.2 cm = 0.052 m

charge of the small plastic bead, q = -9 nC = - 9 x 10⁻⁹ C

The strength of the electric field is calculated as;

[tex]E = \frac{kq}{d^2}[/tex]

where;

E is the electric field strength

k is coulomb's constant = 8.99 x 10⁹ Nm²/C²

[tex]E = \frac{(8.99*10^9)(9*10^{-9})}{(0.052)^2} \\\\E = 29922.34 \ N/C[/tex]

Therefore, the strength of the electric field is 29922.34 N/C

Lanuel

The strength of this electric field is equal to 29,922.34 N/C.

Given the following data:

  • Radius = 5.2 cm
  • Charge = -9.0 nC = [tex]-9 \times 10^{-9}\;C[/tex]

Conversion:

100 cm = 1 m

5.2 cm = 0.052 m

Scientific data:

  • Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]

To calculate the strength of this electric field:

The formula for electric field strength.

Mathematically,  the strength of this electric field is given by this formula:

[tex]E=\frac{qk}{r^2}[/tex]

Where:

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

Substituting the given parameters into the formula, we have;

[tex]E=\frac{-9 \times 10^{-9}\times 8.99 \times 10^9}{0.052^2} \\\\E=\frac{80.91}{0.002704}[/tex]

E = 29,922.34 N/C.

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