Four charges on a rectangle that is 6 cm by 3 cm. All but one of the corners has a positive charge. The magnitudes of the charges are all 8.40e-12 C. What is the net electric field at the center of the rectangle?
The electric field contribution of the two opposite positive charges cancel out. The two charges on opposite sides that are of opposite sign will double the field due to a single charge. The center of the rectangle is at a distance R from corners that is given by the equation
R^2 = 3^2 + 1.5^2
R = 3.354 cm
Q = 8.40*10^-12 C
E = 2 * kQ^2/R^2
To find the net electric field at the center of the rectangle, we need to calculate the individual electric fields due to each charge and then sum them up vectorially.
The formula to calculate the electric field due to a point charge at a specific point is given by:
E = k * (Q / r^2)
Where:
E is the electric field
k is the Coulomb's constant (8.988 × 10^9 N m^2 / C^2)
Q is the charge
r is the distance from the charge to the point where we want to calculate the electric field.
Since the magnitudes of all the charges are the same, let's calculate the electric field due to one charge at the center of the rectangle:
First, let's calculate the distance from the center to one of the charges on the diagonally opposite corner of the rectangle. The diagonal of a rectangle can be calculated using the Pythagorean theorem:
d = √(L^2 + W^2)
Where:
d is the diagonal
L is the length of the rectangle
W is the width of the rectangle
Substituting the values:
d = √(6^2 + 3^2) = √45 = 6.71 cm
Now, let's calculate the electric field due to one charge at the center:
E1 = k * (Q / d^2) = (8.988 × 10^9 N m^2 / C^2) * (8.40 × 10^-12 C / (6.71 × 10^-2 m)^2)
E1 ≈ 1.85 × 10^9 N/C
Next, let's calculate the distance from the center to the charges located on the shorter side of the rectangle. This distance would be half the length of the rectangle, so it is 3 cm or 3 × 10^-2 m.
Now, let's calculate the electric field due to one of these charges at the center:
E2 = k * (Q / r^2) = (8.988 × 10^9 N m^2 / C^2) * (8.40 × 10^-12 C / (3 × 10^-2 m)^2)
E2 ≈ 1.71 × 10^12 N/C
Finally, since we have two charges on the shorter side of the rectangle, the net electric field on the center of the rectangle due to these two charges would be 2 times E2 or 2 × 1.71 × 10^12 N/C.
Therefore, the net electric field at the center of the rectangle would be the vector sum of E1 and 2E2:
Net Electric Field = E1 + 2E2
Substituting the values:
Net Electric Field ≈ 1.85 × 10^9 N/C + 2 × 1.71 × 10^12 N/C
Net Electric Field ≈ 3.34 × 10^12 N/C
Hence, the net electric field at the center of the rectangle is approximately 3.34 × 10^12 N/C.