A charge of -17 nC is on the X axis at the origin. A charge of -24 nC is on the X axis at 11 cm. Find the electric field (magnitude and direction) at 19 cm on the X axis.
I believe to solve this problem, the equation E = kC (q1/r1^2) must be used for each of the charges somehow, but I'm not sure how that would look. Thanks!
E=-k(17/x^2 + 24/(x-11)^2)
x=19
solve for E
To find the electric field at a specific point, you can use the principle of superposition. The electric field at any point on the X axis is the vector sum of the electric fields produced by each charge at that point.
First, let's calculate the electric field produced by the charge of -17 nC at the origin.
Given:
Charge 1, q1 = -17 nC
Distance from the origin to the point, r1 = 19 cm
The electric field, E1, due to Charge 1 can be calculated using the equation:
E1 = (kC * |q1|) / r1^2
where k is the electrostatic constant and C is the unit conversion constant.
Plugging in the values, we get:
E1 = (9 * 10^9 Nm^2/C^2 * 1 C * |-17 * 10^-9 C|) / (0.19 m)^2
Simplifying, we have:
E1 = (9 * 17) / 0.19^2
Next, let's calculate the electric field produced by the charge of -24 nC at 11 cm.
Given:
Charge 2, q2 = -24 nC
Distance from the charge 2 to the point, r2 = 19 cm - 11 cm = 8 cm
Again, using the equation for the electric field:
E2 = (kC * |q2|) / r2^2
Plugging in the values, we get:
E2 = (9 * 10^9 Nm^2/C^2 * 1 C * |-24 * 10^-9 C|) / (0.08 m)^2
Simplifying, we have:
E2 = (9 * 24) / 0.08^2
Now, to find the net electric field, we need to add the two electric fields together because electric fields are vectors.
The net electric field, E_net, at the point on the X axis is given by:
E_net = E1 + E2
After calculating the values for E1 and E2, simply add them to get the magnitude and direction of the net electric field at the point 19 cm on the X axis.