Question

A point charge q1=+2.00 nC is held stationary at the origin.A second point charge q2=-4.50 nC moves from the point xi=0.300 m,yi=0.400 to the point xf=0.600 m,yf=0.800 m.How much work is done by the electric force on q2?

Answer 1

Δx = 0,600 - 0,300 = 0,300 м
Δy = 0,800 - 0,400 = 0,400 м
r = √ (0,300² + 0,400²) = 0,500 м
| F | = | k·q₁·(-q₂) / r² | = 9·10⁹·2·10⁻⁹·4,5·10⁻⁹ / 0,500² = 3,24·10⁻⁷ H
A = F·r = 3,24·10⁻⁷·0,500 =  1,62·10⁻⁷ Дж

Answer 2

To calculate the work done by the electric force on q2, we can use the formula for the work done by a conservative force, which in this case is the electric force:

$\text{Work} = \Delta PE = PE_{\text{final}} - PE_{\text{initial}}$

where $\Delta PE$ is the change in potential energy of charge q2.

The potential energy of a point charge q in the presence of another point charge Q can be given as:

$PE = \frac{k \cdot |q_1 \cdot q_2|}{r}$

where:

• $k$ is 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 two point charges
• $r$ is the distance between the two charges

Since q1 is held stationary, its position does not change, and hence its potential energy does not change. Therefore, $PE_{\text{initial}} = PE_{\text{final}}$ for q1.

Now, let's calculate the potential energy for q2 at the initial position (xi, yi) and the final position (xf, yf).

Initial position: (xi, yi) = (0.300 m, 0.400 m) Final position: (xf, yf) = (0.600 m, 0.800 m)

The distance between q1 and q2 at the initial position is:

$r_{\text{initial}} = \sqrt{(x_i - 0)^2 + (y_i - 0)^2} = \sqrt{0.300^2 + 0.400^2} \approx 0.500 \, \text{m}$

The distance between q1 and q2 at the final position is:

$r_{\text{final}} = \sqrt{(x_f - 0)^2 + (y_f - 0)^2} = \sqrt{0.600^2 + 0.800^2} \approx 1.000 \, \text{m}$

Now, we can calculate the potential energies:

$PE_{\text{initial}} = \frac{k \cdot |q_1 \cdot q_2|}{r_{\text{initial}}} = \frac{8.99 \times 10^9 \cdot |2.00 \times 10^{-9} \cdot (-4.50 \times 10^{-9})|}{0.500}$

$PE_{\text{final}} = \frac{k \cdot |q_1 \cdot q_2|}{r_{\text{final}}} = \frac{8.99 \times 10^9 \cdot |2.00 \times 10^{-9} \cdot (-4.50 \times 10^{-9})|}{1.000}$

Now, we can calculate the work done by the electric force:

$\text{Work} = \Delta PE = PE_{\text{final}} - PE_{\text{initial}}$

$\text{Work} = \frac{8.99 \times 10^9 \cdot |2.00 \times 10^{-9} \cdot (-4.50 \times 10^{-9})|}{1.000} - \frac{8.99 \times 10^9 \cdot |2.00 \times 10^{-9} \cdot (-4.50 \times 10^{-9})|}{0.500}$

Now, perform the calculations:

$\text{Work} = \frac{8.99 \times 10^9 \cdot 9.00 \times 10^{-18}}{1.000} - \frac{8.99 \times 10^9 \cdot 9.00 \times 10^{-18}}{0.500}$