Two charged particle are fixed on the x-axis 11 cm apart. The one on the left has a net charge of +25uc and the one on the right has a net charge of -11uc. What is the magnitude of the electric field 10 cm to the right of the negative charge?
How do you solve this problem?
Thanks!
To solve this problem, we can use Coulomb's law to calculate the electric field at the specified location. Coulomb's law states that the magnitude of the electric field (E) created by a point charge is given by the formula:
E = k * |Q| / r^2
where:
- E is the electric field
- k is Coulomb's constant (k = 9 x 10^9 N*m^2/C^2)
- |Q| is the magnitude of the charge
- r is the distance from the charge to the location where the electric field is being measured
Let's go step-by-step to find the magnitude of the electric field 10 cm to the right of the negative charge.
Step 1: Convert the distances to meters
- The charges are placed 11 cm apart, which is equal to 0.11 m.
- The location where the electric field is being measured is 10 cm to the right of the negative charge, which is equal to 0.10 m.
Step 2: Calculate the electric field due to the left positive charge
- The magnitude of the left positive charge is +25 μC, which is equal to 25 x 10^-6 C.
- The distance from the left positive charge to the location where the electric field is being measured is 0.10 m.
Using the formula E = k * |Q| / r^2, we can calculate the electric field due to the left positive charge.
E1 = (9 x 10^9 N*m^2/C^2) * (25 x 10^-6 C) / (0.10 m)^2
Step 3: Calculate the electric field due to the right negative charge
- The magnitude of the right negative charge is -11 μC, which is equal to -11 x 10^-6 C.
- The distance from the right negative charge to the location where the electric field is being measured is 0.11 m.
Using the formula E = k * |Q| / r^2, we can calculate the electric field due to the right negative charge.
E2 = (9 x 10^9 N*m^2/C^2) * (11 x 10^-6 C) / (0.11 m)^2
Step 4: Calculate the net electric field at the measurement location
To find the net electric field at the measurement location, we need to consider the direction of the electric fields due to the positive and negative charges. Since the electric field due to the positive charge points away from it, and the electric field due to the negative charge points towards it, we subtract the magnitudes to find the net electric field.
Net E = E1 - E2
Calculating the values will give you the final result.
To solve this problem, you can use the concept of electric fields created by point charges. The electric field is a vector quantity that describes the force experienced by a positive test charge placed in the electric field.
The formula to calculate the electric field created by a point charge is given by:
E = k * (|q| / r^2)
Where:
E is the electric field,
k is the electrostatic constant (9 x 10^9 Nm^2/C^2),
|q| is the magnitude of the charge, and
r is the distance between the charge and the point where you want to find the field.
Let's break down the steps to solve the problem:
Step 1: Calculate the electric fields created by each point charge separately.
- For the positive charge (left charge): E1 = k * (|q1| / r1^2)
- For the negative charge (right charge): E2 = k * (|q2| / r2^2)
Step 2: Determine the net electric field at the point of interest.
- Since electric fields are vector quantities, the net electric field is the vector sum of the individual electric fields. In this case, since the charges are on the x-axis, you need to consider only the x-components of the electric fields.
- To add two vectors, you can apply the principle of superposition. The net electric field (Ex) is given by the sum of the x-components of the individual electric fields.
- Ex = E1 - E2
Step 3: Substitute the given values into the formula and perform the calculations.
- You are given the charges: |q1| = +25 μC and |q2| = -11 μC
- The distances are given as: r1 = 11 cm and r2 = 10 cm
Step 4: Find the magnitude of the net electric field.
- The magnitude of the net electric field is given by the absolute value of Ex.
Performing the calculations in step 3 will give you the magnitude of the electric field at the point 10 cm to the right of the negative charge.