Two infinitely long, parallel lines of charge with linear charge densities 4.1 µC/m and −4.1 µC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in the figure below? (Express your answers in vector form.)
Enet,A=N/C
Enet,B=N/C
Enet,C=N/C
webassign.net/katzpse1/25-p-035-alt.png
To find the net electric field at points A, B, and C, we need to calculate the electric field contribution due to each line of charge and then add them vectorially.
1. Point A:
At point A, the electric field due to the positively charged line is directed towards the line of charge, and the electric field due to the negatively charged line is directed away from the line of charge. Since both lines have the same linear charge density, their electric field magnitudes will be equal in magnitude but opposite in direction. The net electric field at A is the vector sum of the electric fields due to each line.
Let's first calculate the electric field due to each line at point A:
- The electric field due to the positively charged line can be calculated using the formula: E_1 = (k * λ_1) / r_1, where k is the Coulomb constant, λ_1 is the linear charge density of the positive line, and r_1 is the distance from the positive line to point A.
- Similarly, the electric field due to the negatively charged line can be calculated using the formula: E_2 = (k * λ_2) / r_2, where λ_2 is the linear charge density of the negative line, and r_2 is the distance from the negative line to point A.
Now, we can plug in the values:
- Given: λ_1 = 4.1 µC/m, λ_2 = -4.1 µC/m, r_1 = 0.25 m (half of the given distance), r_2 = 0.25 m (half of the given distance), and k = 9 × 10^9 N m^2/C^2.
Calculating the electric field due to the positive line:
E_1 = (9 × 10^9 N m^2/C^2 * 4.1 × 10^-6 C/m) / 0.25 m
Calculating the electric field due to the negative line:
E_2 = (9 × 10^9 N m^2/C^2 * -4.1 × 10^-6 C/m) / 0.25 m
Next, we sum these two electric fields vectorially to find the net electric field at point A.
2. Point B:
At point B, both lines of charge are equidistant from the point, but since they have opposite linear charge densities, their electric field contributions will be in the same direction. The magnitudes of the electric fields due to each line will simply add up.
Calculate the electric field due to the positive line at point B using the same formula as before, and then calculate the electric field due to the negative line at point B as well. Add both electric fields together to find the net electric field at point B.
3. Point C:
At point C, the distance between the two lines of charge is equal to the given distance of 0.50 m. Since point C is equidistant from both lines, the electric field contributions due to each line will have equal magnitudes but opposite directions, just like at point A.
Calculate the electric field due to the positive line at point C using the same formula as before but with the full distance of 0.50 m as r_1. Then calculate the electric field due to the negative line at point C as well but with the full distance of 0.50 m as r_2. Finally, add both electric fields together to find the net electric field at point C.
Recall that vectors add up according to their components, so make sure to consider the direction and magnitude of each electric field contribution when adding them.