1) Charges of +4uC and -6uC are placed at two corners of an equilateral triangle with sides of 0.10m. At the third corner, what is the electric field magnitude created by these two charges?
A]4.5 x 10e6 N/C
B]3.1 x 10e6 N/C
C]1.6 x 10e6
D]4.8 x 10e6
2)A +2.0 nC pont ahrge isplaced at one corner of a square (1.5m on each side), and a -3.0 nC charge is placed on a corner diagonally away from the first charge. What is the magnitude of the electric fled at either the two unoccupied corners?
A]20 N/C
B]14 N/C
C]4 N/C
D]12 N/C
E]8 N/C
4.8 x 10^6 N/C
1) To find the electric field magnitude at the third corner of the equilateral triangle, we can use the superposition principle.
First, we need to find the electric field created by each individual charge at the third corner.
The electric field created by a point charge can be calculated using the formula:
E = k * q / r^2
Where E is the electric field, k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to calculate the field.
For the +4uC charge:
E1 = (8.99 x 10^9 Nm^2/C^2) * (+4 x 10^-6 C) / (0.10 m)^2
E1 = 3.596 x 10^5 N/C
For the -6uC charge:
E2 = (8.99 x 10^9 Nm^2/C^2) * (-6 x 10^-6 C) / (0.10 m)^2
E2 = -5.394 x 10^5 N/C
Next, we can find the resultant electric field at the third corner by summing the electric fields created by the individual charges. Since the charges have opposite signs, we need to take their difference:
E_total = E1 - E2
E_total = (3.596 x 10^5 N/C) - (-5.394 x 10^5 N/C)
E_total = 8.99 x 10^5 N/C
Therefore, the magnitude of the electric field created by the +4uC and -6uC charges at the third corner of the equilateral triangle is 8.99 x 10^5 N/C.
C] 1.6 x 10^6 N/C
2) To find the magnitude of the electric field at the two unoccupied corners of the square, we can again use the superposition principle.
Since the +2.0 nC charge and the -3.0 nC charge have the same magnitude but opposite signs, the electric field created by each charge at the unoccupied corners will be the same in magnitude, but opposite in direction.
The electric field created by a point charge can be calculated using the formula:
E = k * q / r^2
Where E is the electric field, k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to calculate the field.
For the +2.0 nC charge and the -3.0 nC charge, we can use the same formula to calculate the electric field at either of the unoccupied corners.
E1 = (8.99 x 10^9 Nm^2/C^2) * (+2 x 10^-9 C) / (1.5 m)^2
E1 = 10.662 x 10^6 N/C (approximately)
E2 = (8.99 x 10^9 Nm^2/C^2) * (-3 x 10^-9 C) / (1.5 m)^2
E2 = -15.993 x 10^6 N/C (approximately)
The magnitudes of these electric fields are equal, so the magnitude of the electric field at either of the unoccupied corners is:
|E| = |E1| = |E2| = 10.662 x 10^6 N/C (approximately)
Therefore, the magnitude of the electric field at the two unoccupied corners of the square is approximately 10.662 x 10^6 N/C.
None of the given answer choices match exactly.
To find the electric field magnitude at a point, we need to calculate the electric field contributions from each charge and then add them together. The magnitude of the electric field created by a point charge at a certain distance is given by the equation:
E = k * |q| / r^2
where E is the electric field magnitude, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), |q| is the absolute value of the charge (in Coulombs), and r is the distance between the point charge and the point where we want to calculate the electric field (in meters).
Let's solve the first problem:
1) Charges of +4uC and -6uC are placed at two corners of an equilateral triangle with sides of 0.10m. At the third corner, what is the electric field magnitude created by these two charges?
Step 1: Calculate the electric field contribution from the +4uC charge at the third corner.
E1 = k * |q1| / r^2
E1 = (8.99 x 10^9 N m^2/C^2) * (4 x 10^-6 C) / (0.10 m)^2
Step 2: Calculate the electric field contribution from the -6uC charge at the third corner.
E2 = k * |q2| / r^2
E2 = (8.99 x 10^9 N m^2/C^2) * (6 x 10^-6 C) / (0.10 m)^2
Step 3: Add the electric field contributions together to find the total electric field at the third corner.
E_total = E1 + E2
Plug in the values and calculate the result to find the correct answer.
Let's solve the second problem:
2) A +2.0 nC point charge is placed at one corner of a square (1.5m on each side), and a -3.0 nC charge is placed on a corner diagonally away from the first charge. What is the magnitude of the electric field at either of the two unoccupied corners?
Step 1: Calculate the electric field contribution from the +2.0 nC charge at one of the unoccupied corners.
E1 = k * |q1| / r^2
E1 = (8.99 x 10^9 N m^2/C^2) * (2 x 10^-9 C) / (1.5 m)^2
Step 2: Calculate the electric field contribution from the -3.0 nC charge at one of the unoccupied corners.
E2 = k * |q2| / r^2
E2 = (8.99 x 10^9 N m^2/C^2) * (3 x 10^-9 C) / (1.5 m)^2
Step 3: Add the electric field contributions together to find the total electric field at one of the unoccupied corners.
E_total = E1 + E2
Plug in the values and calculate the result to find the correct answer.
Both problems involve the adding of two vectors. Get the magnitude of each vector using Coulomb's law.
You should make an effort to do these yourself.
In problem 2), you end up adding two perpendicular E-field vectors with magnitudes
k*2*10^-9/(1.5)^2 and
k*3*10^-9/(1.5)^2
The magnitude is, from the Pythagorean Theorem,
(k/2.25)*10^-9 sqrt13
Using 9.0*10^9 for k, I get 14.4 N/C