Question

An electron with a charge of -1.6 x 10-19C is placed 0.22 meters from a small metal sphere that is given a charge of 6.7 x 10-4C. What is the magnitude of the electrical force acting on the charges?

Answer 1

To calculate the magnitude of the electrical force between the electron and the charged metal sphere, we can use Coulomb's law, which is given by the formula:

\[
F = k \frac{|q_1 \cdot q_2|}{r^2}
\]

where:
- \( F \) is the magnitude of the electrical force between the charges,
- \( k \) is the electrostatic constant (Coulomb's constant), approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

In this case:
- \( q_1 = -1.6 \times 10^{-19} \, \text{C} \) (charge of the electron),
- \( q_2 = 6.7 \times 10^{-4} \, \text{C} \) (charge of the metal sphere),
- \( r = 0.22 \, \text{m} \).

First, we will calculate the absolute values of the charges:

\[
|q_1| = 1.6 \times 10^{-19} \, \text{C}
\]
\[
|q_2| = 6.7 \times 10^{-4} \, \text{C}
\]

Now, substituting the values into Coulomb's law equation:

\[
F = (8.99 \times 10^9) \frac{(1.6 \times 10^{-19})(6.7 \times 10^{-4})}{(0.22)^2}
\]

Calculating \( (0.22)^2 \):

\[
(0.22)^2 = 0.0484 \, \text{m}^2
\]

Now, substituting this back into the equation:

\[
F = (8.99 \times 10^9) \frac{(1.6 \times 10^{-19})(6.7 \times 10^{-4})}{0.0484}
\]

Calculating the numerator:

\[
(1.6 \times 10^{-19}) \times (6.7 \times 10^{-4}) = 1.072 \times 10^{-22} \, \text{C}^2
\]

Now putting this into the force equation:

\[
F = (8.99 \times 10^9) \frac{1.072 \times 10^{-22}}{0.0484}
\]

Calculating the fraction:

\[
\frac{1.072 \times 10^{-22}}{0.0484} = 2.21 \times 10^{-21}
\]

Now substituting it back:

\[
F = (8.99 \times 10^9) \times (2.21 \times 10^{-21}) \approx 0.01988 \, \text{N}
\]

Therefore, the magnitude of the electrical force acting on the charges is approximately:

\[
\boxed{0.0199 \, \text{N}}
\]