Two charges are placed on the x axis. One of the charges (q1 = +8.59C) is at x1 = +3.00 cm and the other (q2 = -25.6C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.
(a) Number
Units
To find the net electric field at a certain point, we need to calculate the electric field contribution from each individual charge and then add them up.
The electric field created by a point charge can be found using Coulomb's law:
E = k * |q| / r^2
Where:
- E is the electric field
- k is the electrostatic constant (9.0 * 10^9 N m^2/C^2)
- |q| is the magnitude of the charge
- r is the distance between the point charge and the observation point
(a) Calculating the net electric field at x = 0 cm:
We have two charges, q1 = +8.59C at x1 = +3.00 cm, and q2 = -25.6C at x2 = +9.00 cm.
The distance between the observation point (x = 0 cm) and the first charge (x1 = +3.00 cm) is:
r1 = |x - x1| = |0 - 3.00| = 3.00 cm
The electric field contributed by q1 at x = 0 cm is:
E1 = k * |q1| / r1^2
The distance between the observation point and the second charge (x2 = +9.00 cm) is:
r2 = |x - x2| = |0 - 9.00| = 9.00 cm
The electric field contributed by q2 at x = 0 cm is:
E2 = k * |q2| / r2^2
The net electric field at x = 0 cm is the vector sum of E1 and E2:
E_net = E1 + E2
Now, we can plug in the values and calculate:
E1 = (9.0 * 10^9 N m^2/C^2) * (8.59C) / (3.00 cm)^2
E2 = (9.0 * 10^9 N m^2/C^2) * (-25.6C) / (9.00 cm)^2
Add the magnitudes of E1 and E2 to get the net electric field's magnitude at x = 0 cm. The direction will be positive or negative depending on the sign of the result.
(b) Calculating the net electric field at x = +6.00 cm:
We follow the same procedure as above, but now the observation point is at x = +6.00 cm.
The distance between the observation point (x = +6.00 cm) and the first charge (x1 = +3.00 cm) is:
r1 = |x - x1| = |6.00 - 3.00| = 3.00 cm
The electric field contributed by q1 at x = +6.00 cm is:
E1 = k * |q1| / r1^2
The distance between the observation point and the second charge (x2 = +9.00 cm) is:
r2 = |x - x2| = |6.00 - 9.00| = 3.00 cm
The electric field contributed by q2 at x = +6.00 cm is:
E2 = k * |q2| / r2^2
The net electric field at x = +6.00 cm is the vector sum of E1 and E2:
E_net = E1 + E2
Again, plug in the values and calculate:
E1 = (9.0 * 10^9 N m^2/C^2) * (8.59C) / (3.00 cm)^2
E2 = (9.0 * 10^9 N m^2/C^2) * (-25.6C) / (3.00 cm)^2
Add the magnitudes of E1 and E2 to get the net electric field's magnitude at x = +6.00 cm. The direction will be positive or negative depending on the sign of the result.