1. A charge (uniform linear density=9.0 nC/m) is distributed along the x-axis from x=0 to x=3.0m. Determine the magnitude of the electric field at a point on the x axis with x=4.0m.
a)81 N/C
b)74 N/C
c)61 N/C
d)88 N/C
e)20 N/C
2. A +2nC point charge is placed at one corner of a square (1.5m on a side), and a -3.0nC 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?
a)20 N/C
b)14 N/C
c)4 N/C
d)12 N/C
e)8 N/C
1. To determine the magnitude of the electric field at a point on the x-axis with x=4.0m, we need to calculate the electric field contribution from each section of the charge distribution and then add them up.
First, let's consider the charge distribution from x=0 to x=3.0m. The linear charge density is given as 9.0 nC/m. We can calculate the charge contained within this length by multiplying the linear charge density by the length:
Charge = (linear charge density) * (length)
Charge = (9.0 nC/m) * (3.0m)
Charge = 27 nC
Next, we can calculate the electric field contribution at x=4.0m due to this charge. The electric field due to a point charge is given by Coulomb's Law:
Electric field (E) = (k * q) / r^2
where k is the electrostatic constant (9.0 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.
In this case, q = 27 nC and r = 4.0m. Plugging these values into the formula, we get:
E = (9.0 x 10^9 Nm^2/C^2) * (27 x 10^-9 C) / (4.0m)^2
Solving this equation gives us:
E = 81 N/C.
Therefore, the magnitude of the electric field at x=4.0m is 81 N/C.
The correct answer is (a) 81 N/C.
2. To determine the magnitude of the electric field at either of the two unoccupied corners of the square, we can calculate the electric field contributions from each charge separately and then vectorially add them.
Let's consider the +2nC charge at one corner first. The electric field at a point due to a point charge is given by Coulomb's Law:
Electric field (E) = (k * q) / r^2
where k is the electrostatic constant, q is the charge, and r is the distance from the charge to the point.
In this case, q = +2nC and r is the distance from the corner to the unoccupied corner of the square. Since the square has sides of length 1.5m, the distance r is equal to 1.5m * sqrt(2).
Plugging in the values into the formula, we get:
E1 = (9.0 x 10^9 Nm^2/C^2) * (2 x 10^-9 C) / (1.5m * sqrt(2))^2
Solving this equation gives us:
E1 = 14 N/C.
Now let's consider the -3.0nC charge at the other corner. Using the same formula, we get:
E2 = (9.0 x 10^9 Nm^2/C^2) * (-3 x 10^-9 C) / (1.5m * sqrt(2))^2
Solving this equation gives us:
E2 = -14 N/C.
The negative sign indicates that the electric field strictly points away from the charge.
Since the electric fields due to the two charges have the same magnitude but opposite directions, we can add them vectorially to find the net electric field at the unoccupied corner:
Net electric field = E1 + E2 = (14 N/C) + (-14 N/C) = 0 N/C.
Therefore, the magnitude of the electric field at either of the two unoccupied corners is 0 N/C.
The correct answer is (c) 0 N/C.
To determine the magnitude of the electric field at a certain point, we need to use the concept of the electric field due to a continuous charge distribution and the principle of superposition.
1. For the first question, we have a charge distributed along the x-axis from x = 0 to x = 3.0m with a uniform linear density of 9.0 nC/m. We can calculate the electric field at a point on the x-axis with x = 4.0m using the following steps:
Step 1: Divide the charge distribution into small elements:
Divide the charge distribution into small elements, each with a length Δx. The smaller the length of each element, the more accurate the result will be. Let's assume Δx = 0.01m for this calculation.
Step 2: Determine the electric field contribution due to each element:
For each small element, calculate the electric field at the point using Coulomb's law.
The electric field contribution due to each small element, dE, is given by dE = (k * dq) / r^2, where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2), dq is the charge of the small element, and r is the distance between the small element and the point where we want to find the electric field.
Step 3: Integrate the contributions from all small elements:
To find the electric field at the point, sum up the electric field contributions from each small element. Integrate the individual dE values from 0 to 3.0m to find the total electric field at x = 4.0m.
Step 4: Calculate the magnitude of the electric field:
The magnitude of the electric field is given by the absolute value of the total electric field calculated in step 3.
By following these steps, the magnitude of the electric field at x = 4.0m can be determined.
2. For the second question, we have two point charges placed at the diagonally opposite corners of a square. One charge has a value of +2nC and the other has a value of -3.0nC. We need to calculate the magnitude of the electric field at either of the two unoccupied corners of the square.
Step 1: Find the electric field contribution due to each charge:
Using Coulomb's law, calculate the electric field at one of the unoccupied corners caused by the +2nC charge and the -3.0nC charge separately. The electric field due to a point charge is given by E = (k * q) / r^2, where k is the electrostatic constant, q is the charge, and r is the distance between the charge and the point where we want to find the electric field.
Step 2: Find the total electric field at the unoccupied corner:
Since the electric field is a vector quantity, we need to find the vector sum of the electric fields due to the +2nC and -3.0nC charges. Both electric fields will have a direction, so we need to add them as vectors, considering both magnitude and direction.
Step 3: Calculate the magnitude of the electric field:
The magnitude of the electric field is given by the absolute value of the vector sum calculated in step 2. This magnitude represents the strength of the electric field at either of the two unoccupied corners.
By following these steps, the magnitude of the electric field at the unoccupied corner of the square can be determined.
2. What is Pythagoras theorem? figure the E from each, then use the theorem.
1. Simple integration. draw the figure.
E=int k (9.0E-9)/(4-x)^2 dx
integrate from x=0 to 3