answer key: -6.40*10^13i N/C Can somebody show me the steps to the answer key?
The figure below shows a dipole. If the positive particle has a charge of 37.1 mC and the particles are 2.66 mm apart, what is the electric field at point A located 2.00 mm above the dipole's midpoint? (Express your answer in vector form.)
The figure is a xy plane with A on the positive y axis. The negative charge is on the left of the x axis and positive charge is on the right of the y axis.
To find the electric field at point A, we can use the principle of superposition, which states that the total electric field is the vector sum of the individual electric fields produced by each charge in the system.
Here are the step-by-step calculations:
1. Determine the electric field generated by the positive charge at point A:
The formula for the electric field generated by a point charge is given by:
E = k * q / r^2, where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point where the field is being measured.
Plugging in the values:
q = 37.1 mC = 37.1 x 10^(-3) C
r = 2.00 mm = 2.00 x 10^(-3) m
Calculating the electric field generated by the positive charge:
E1 = (8.99 x 10^9 N m^2/C^2) * (37.1 x 10^(-3) C) / (2.00 x 10^(-3) m)^2
2. Determine the electric field generated by the negative charge at point A:
Since the positive charge is located above the negative charge, the electric field generated by the negative charge will point upwards (opposite to the positive y-axis).
Using the same formula and values as above, we get:
E2 = (8.99 x 10^9 N m^2/C^2) * (-37.1 x 10^(-3) C) / (2.00 x 10^(-3) m)^2
3. Add the electric fields generated by the positive and negative charges:
Since the electric field is a vector quantity, we need to sum the magnitudes and consider their directions.
The electric field generated by the positive charge points downwards (along the positive y-axis), while the electric field generated by the negative charge points upwards (opposite to the positive y-axis).
Adding the magnitudes, we have:
E_total = |E1| + |E2|
Since |E1| = |E2|, we can simplify the equation:
E_total = 2 * |E1|
Evaluating the magnitude:
E_total = 2 * [ (8.99 x 10^9 N m^2/C^2) * (37.1 x 10^(-3) C) / (2.00 x 10^(-3) m)^2 ]
4. Expressing the answer in vector form:
The electric field points downwards, along the positive y-axis, so its direction can be represented in vector form as:
E_total = 0i - E_total j,
where "i" and "j" represent the unit vectors along the x-axis and y-axis, respectively.
Therefore, the final answer in vector form is:
E_total = -6.40 x 10^13 N/C j