What is the electric field at point P due to q1 and q2?

Steps to Solve

1. Identify the Charges and Their Positions

   We have two charges:

   • $q_1 = 3.00 , \text{nC}$ located at position $(0, 0) , \text{m}$
   • $q_2 = 4.00 , \text{nC}$ located at position $(1, 0) , \text{m}$ The point P is located at $(0.5, 0) , \text{m}$.

2. Calculate the Distances from Charges to Point P

   The distance from $q_1$ to point P is: $$ d_1 = 0.5 , \text{m} $$ The distance from $q_2$ to point P is: $$ d_2 = 0.5 , \text{m} $$

3. Calculate the Electric Field Contributions

   The electric field due to a point charge can be calculated using the formula: $$ E = \frac{k \cdot |q|}{r^2} $$ Where ( k \approx 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 ).

   For $q_1$: $$ E_1 = \frac{8.99 \times 10^9 \cdot |3.00 \times 10^{-9}|}{(0.5)^2} $$

   For $q_2$: $$ E_2 = \frac{8.99 \times 10^9 \cdot |4.00 \times 10^{-9}|}{(0.5)^2} $$

4. Determine the Direction of Each Electric Field

   The electric field due to $q_1$ at point P will point away from $q_1$ (to the right), and the electric field due to $q_2$ will point toward $q_2$ (to the left).

5. Calculate the Magnitudes of the Electric Fields

   Compute the values: $$ E_1 = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{0.25} $$ $$ E_2 = \frac{8.99 \times 10^9 \cdot 4.00 \times 10^{-9}}{0.25} $$

6. Combine the Electric Fields

   Since $E_1$ acts to the right and $E_2$ acts to the left, the net electric field at point P is: $$ E_{\text{net}} = E_1 - E_2 $$

7. Calculate the Final Electric Field Value

   Substitute the calculated values of $E_1$ and $E_2$ into the equation to find $E_{\text{net}}$.