Calculating Work Done Moving Electric Charge to Point P

This problem involves calculating the work done required to move an electric charge from an infinite distance to a specific point (P) in space, where another charge is already present. The work done in such a scenario is directly related to the electric potential at that point and the magnitude of the charge being moved.

Understanding Electric Potential and Work Done

The electric potential ($V$) at any point in space due to a point charge ($q_1$) is the amount of work needed per unit charge to move a test charge from infinity to that point. The formula for electric potential at a distance ($r$) from a point charge ($q_1$) is given by:

$V = k \frac{q_1}{r}$

where $k$ is Coulomb's constant, approximately $9 \times 10^9$ N m²/C².

The work done ($W$) in bringing another charge ($q_2$) from infinity to this point P is the product of the charge $q_2$ and the electric potential $V$ at point P:

$W = q_2 \times V$

Substituting the expression for $V$, we get the formula for work done:

$W = k \frac{q_1 q_2}{r}$

Given Values

• Magnitude of the first charge ($q_1$): $3 \times 10^{-7}$ C
• Magnitude of the second charge ($q_2$): $2 \times 10^{-9}$ C
• Distance from the first charge to point P ($r$): 0.09 m
• Coulomb's constant ($k$): $9 \times 10^9$ N m²/C²

Step-by-Step Calculation

To find the work done, we will substitute the given values into the work done formula:

1. Write down the formula: The work done ($W$) is calculated using the formula $W = k \frac{q_1 q_2}{r}$.
2. Substitute the values: Plug in the values for $k$, $q_1$, $q_2$, and $r$ into the formula.

   $W = (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \times \frac{(3 \times 10^{-7} \, \text{C}) \times (2 \times 10^{-9} \, \text{C})}{0.09 \, \text{m}}$

3. Calculate the product of charges: Multiply the two charges together.

   $q_1 \times q_2 = (3 \times 10^{-7}) \times (2 \times 10^{-9}) = 6 \times 10^{-16} \, \text{C}^2$

4. Calculate the numerator: Multiply the result from step 3 by Coulomb's constant ($k$).

   $k \times q_1 \times q_2 = (9 \times 10^9) \times (6 \times 10^{-16}) = 54 \times 10^{-7} \, \text{N m}^2/\text{C}$

5. Divide by the distance: Divide the result from step 4 by the distance ($r$).

   $W = \frac{54 \times 10^{-7} \, \text{N m}^2/\text{C}}{0.09 \, \text{m}}$

   To simplify the division, we can write 0.09 as $9 \times 10^{-2}$.

   $W = \frac{54 \times 10^{-7}}{9 \times 10^{-2}} \, \text{N m}/\text{C}$

   $W = \left(\frac{54}{9}\right) \times 10^{(-7 - (-2))} \, \text{J}$

   $W = 6 \times 10^{-7 + 2} \, \text{J}$

   $W = 6 \times 10^{-5} \, \text{J}$

Therefore, the work done in bringing the charge of $2 \times 10^{-9}$ C from infinity to the point P is $6 \times 10^{-5}$ Joules.