Three 3.0 microC charges are at the three corners of an square of side 0.50 m. The last corner is occupied by a -3.0 microC charge. Find the electric field at the center of the square
Use the principle of superposition. The E field at the center will be the sum of two cases:
(a) +3 microC at the open corner, and all three other corners
(b) -6 microC at the open corner location only
Case (a) results in a net E field at the center of zero. That leaves you with case (b), which is easily solved.
584km
The total E is zero
410n/c
To find the electric field at the center of the square, we need to calculate the electric field due to each individual charge and then sum up the vectors.
Step 1: Calculate the electric field due to a single charge.
The magnitude of the electric field due to a point charge can be calculated using the formula:
E = k * (Q / r^2)
where E is the electric field, k is the 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.
Let's assume the side length of the square is d. The distance from the center of the square to any of the charges is d/2.
For the positive charges at the corners, the electric field vectors point away from the charges. Thus, each positive charge contributes an electric field vector pointing away from itself.
For the negative charge at the last corner, the electric field vector points towards the charge. So, the electric field due to the negative charge will have a direction opposite to that of the positive charges.
Step 2: Calculate the electric field due to each individual charge at the center.
To find the net electric field at the center, we need to consider the electric field vectors from all four charges.
The total electric field vector at the center will be the vector sum of the individual electric field vectors.
Step 3: Calculate the total electric field at the center.
Since the electric field is a vector quantity, we need to consider both the magnitude and direction of each electric field vector.
The magnitude of the total electric field is the sum of the magnitudes of the individual electric fields.
The direction of the total electric field can be determined by applying vector addition rules.
Let's do the calculations:
1. Magnitude of the electric field due to a single charge:
E = k * (Q / r^2)
E = (8.99 x 10^9 N m^2/C^2) * (3.0 x 10^-6 C) / ((0.5/2)^2)
E = 8.99 x 10^9 N m^2/C^2 * 3.0 x 10^-6 C / (0.25 m^2)
E = 8.99 x 3.0 x 10^3 N / m
2. Electric field at the center due to the positive charges:
The electric fields from the positive charges will have the same magnitude since the charges are identical.
The vector sum of the two electric field vectors from the positive charges will form an angle of 90 degrees, pointing diagonally towards the center of the square.
Thus, the magnitude of the electric field at the center due to the positive charges will be:
E_positive = 2 * E = 2 * 8.99 x 3.0 x 10^3 N / m
3. Electric field at the center due to the negative charge:
The electric field from the negative charge will have the same magnitude but opposite direction compared to the positive charges.
Thus, the magnitude of the electric field at the center due to the negative charge will be:
E_negative = E = 8.99 x 3.0 x 10^3 N / m
4. Total electric field at the center:
The total electric field vector at the center will be the vector sum of the individual electric field vectors.
Since the electric fields from the positive charges form a 90-degree angle, they can be treated as perpendicular vectors.
The magnitude of the total electric field (E_total) will be the square root of the sum of the squares of the magnitudes:
E_total = sqrt((E_positive)^2 + (E_positive)^2 + (E_negative)^2)
The direction of the total electric field can be determined using vector addition.