Two charges are placed on the x axis. One of the charges (q1 = +9.3 µC) is at x1 = +3.4 cm and the other (q2 = -20 µC) is at x2 = +8.9 cm.
(a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
(b) Find the net electric field (magnitude and direction) at x = +5.4 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
Both answers in (N/C)
To find the net electric field at a certain point on the x-axis due to two charges, we can use the principle of superposition. The net electric field at any point is the vector sum of the electric fields produced by each charge individually.
The electric field produced by a point charge is given by Coulomb's Law. The magnitude of the electric field produced by a charge q at a distance r from it is given by:
E = k * |q| / r^2,
where k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2).
(a) To find the net electric field at x = 0 cm, we need to calculate the electric fields produced by each charge and add them up. The direction of the electric field is indicated by the sign of the charge.
For the charge q1 = +9.3 µC at x1 = +3.4 cm, the distance from it to x = 0 cm is r = |x1 - x| = |3.4 cm - 0 cm| = 3.4 cm. Convert it to meters: r = 3.4 cm * (1 m / 100 cm) = 0.034 m.
Using Coulomb's Law, the electric field produced by q1 at x = 0 cm is:
E1 = k * |q1| / r^2 = (8.99 x 10^9 N m^2/C^2) * (9.3 x 10^-6 C) / (0.034 m)^2.
Calculate E1 to find its magnitude and direction.
For the charge q2 = -20 µC at x2 = +8.9 cm, the distance from it to x = 0 cm is r = |x2 - x| = |8.9 cm - 0 cm| = 8.9 cm. Convert it to meters: r = 8.9 cm * (1 m / 100 cm) = 0.089 m.
Using Coulomb's Law, the electric field produced by q2 at x = 0 cm is:
E2 = k * |q2| / r^2 = (8.99 x 10^9 N m^2/C^2) * (20 x 10^-6 C) / (0.089 m)^2.
Calculate E2 to find its magnitude and direction.
The net electric field at x = 0 cm is the vector sum of E1 and E2. Take into account the signs and add the magnitudes to find the net electric field's magnitude. The direction of the net electric field is along the x-axis and is determined by the sign of the net electric field.
(b) The process for finding the net electric field at x = +5.4 cm is the same as in part (a). Repeat the calculations for E1 and E2 using the new values for the distances. Add the magnitudes, consider the signs, and determine the direction along the x-axis.
Perform the necessary calculations to find the magnitudes and directions of the net electric fields at x = 0 cm and x = +5.4 cm.