What are the magnitude and direction of the electric field at a point midway between a -7.0 µC and a +6.0 µC charge 2.5 cm apart? Assume no other charges are nearby.
At the midway point, the distance is r = 0.0125 m from each charge.
The fields due to each charge are in the same direction at points in between.
Coulomb's law tells you that the E-field due to a charge q is
E = k q/r^2
Add the fields of each charge. You may need to look up the Coulomb constan,t k.
a) What is the magnitude of the electric field at a point midway between a −8.5μC and a +8.5μC charge 8.0cm apart? Assume no other charges are nearby.
b) What is the direction of the electric field?
To find the magnitude and direction of the electric field at a point midway between two charges, we can use the principle of superposition. The electric field at this point is the vector sum of the electric fields produced by each charge individually.
Let's assign a positive direction for the electric field pointing towards the positive charge (+6.0 µC) and a negative direction for the electric field pointing towards the negative charge (-7.0 µC).
First, let's calculate the electric field at the point due to the +6.0 µC charge:
Given:
Charge of the positive charge (q1) = +6.0 µC
Distance from the positive charge to the point (r1) = 1.25 cm (since the point is midway between the charges)
The electric field produced by a point charge is given by the formula:
E1 = k*q1 / r1^2
where k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2).
Substituting the given values, we have:
E1 = (8.99 x 10^9 N m^2/C^2) * (6.0 x 10^-6 C) / (0.0125 m)^2
Calculating this, we find:
E1 = 8628 N/C
Next, let's calculate the electric field at the point due to the -7.0 µC charge:
Given:
Charge of the negative charge (q2) = -7.0 µC
Distance from the negative charge to the point (r2) = 1.25 cm (since the point is midway between the charges)
Using the same formula, we can calculate E2:
E2 = k*q2 / r2^2
Substituting the given values, we have:
E2 = (8.99 x 10^9 N m^2/C^2) * (-7.0 x 10^-6 C) / (0.0125 m)^2
Calculating this, we find:
E2 = -12040 N/C
Now, the total electric field at the midpoint is the vector sum of E1 and E2:
E_total = E1 + E2
Substituting the calculated values, we get:
E_total = 8628 N/C + (-12040 N/C)
Calculating this, we find:
E_total = -3412 N/C
Therefore, the magnitude of the electric field at the midpoint is 3412 N/C, and the direction of the electric field is negative, indicating that it points towards the negative charge.
To find the magnitude and direction of the electric field at the point midway between two charges, you can use Coulomb's Law and the principle of superposition.
1. Coulomb's Law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It can be written as:
F = k * |q1 * q2| / r^2
where F is the electric force, k is the electrostatic constant (8.99 x 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
2. Consider the two charges, -7.0 µC and +6.0 µC, separated by a distance of 2.5 cm (which is 0.025 m). The charges have opposite signs, so the force between them will be attractive.
3. First, calculate the magnitude of the force between the charges using Coulomb's Law:
F = (8.99 x 10^9 N·m^2/C^2) * |-7.0 x 10^(-6) C * 6.0 x 10^(-6) C| / (0.025 m)^2
F = 2.495 x 10^(-3) N
4. Since the force between the charges is attractive, the direction of the electric field at the midpoint will be along the line connecting the charges, from the positive charge (+6.0 µC) towards the negative charge (-7.0 µC).
5. To find the electric field strength at the midpoint, divide the magnitude of the force by the charge placed at that point. Since the charges are the same, we need to consider the net charge at the midpoint:
q_net = q1 + q2
q_net = (-7.0 x 10^(-6) C) + (6.0 x 10^(-6) C)
q_net = -1.0 x 10^(-6) C
6. Now, calculate the electric field strength:
E = F / q_net
E = (2.495 x 10^(-3) N) / (-1.0 x 10^(-6) C)
E = -2.495 x 10^6 N/C
7. The magnitude of the electric field at the midpoint is 2.495 x 10^6 N/C (since electric field is a vector quantity, we take the magnitude).
8. The direction of the electric field is negative, meaning it points from the positive charge towards the negative charge.
Therefore, the magnitude of the electric field at the point midway between the -7.0 µC and +6.0 µC charge is 2.495 x 10^6 N/C, and it points from the positive charge towards the negative charge.