Two charges are placed on the x axis. One of the charges (q1 = +9.4 µC) is at x1 = +3.1 cm and the other (q2 = -22 µC) is at x2 = +9.3 cm.
Find the net electric field (magnitude and direction) at x = +6.5 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
I changed the radius by 6.50-3.1 and then 6.5-9.3 and but then in the equation and then added but im not getting the right answer.
Both E are in the direction to the right. So they add.
Etotal= kqleft/(.065-.031)^2 + kqright/(.094-065)^2
You can me fancy to let the sign do the direction, as in here...
E=kq1 *(.065-.031)/ABS(.065-.031)^3 + kq2(.065-.093)/ABS(.065-.093)^3
That again, will yield a E that is +,to the right.
To find the net electric field at a given point, you need to consider the contributions from both charges and add them vectorially. The electric field at a point due to a single charge is given by Coulomb's Law:
E = k * q / r^2
Where:
- E is the electric field
- k is the electrostatic constant (k ≈ 9 × 10^9 N m^2/C^2)
- q is the charge
- r is the distance between the charge and the point where you want to find the electric field
Let's calculate the electric fields due to both charges at the point x = 6.5 cm:
For q1 (9.4 µC) at x1 (3.1 cm):
r1 = 6.5 cm - 3.1 cm = 3.4 cm = 0.034 m
E1 = k * q1 / r1^2
For q2 (-22 µC) at x2 (9.3 cm):
r2 = 6.5 cm - 9.3 cm = -2.8 cm = -0.028 m (negative because it is in the opposite direction)
E2 = k * q2 / r2^2
Now, you just need to calculate the magnitudes and directions of both electric fields, and then add them together to get the net electric field:
Magnitude of E1 = k * |q1| / r1^2
Magnitude of E2 = k * |q2| / r2^2
To find the direction, you can consider that the electric field due to q1 points away from it, while the electric field due to q2 points towards it (since it is a negative charge).
Finally, to get the net electric field, add the magnitudes of both electric fields and take into account their directions. If the net electric field is positive, it points in the positive x-direction. If it is negative, it points in the negative x-direction.
Let's calculate the individual electric fields and then add them to find the net electric field.