Two charged particles 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 -15uc. What is the magnitude of the electric field 15 cm to the right of the negative charge?
How do you solve this problem?
Thanks!
you can add them:
E=k(+25E-6/.26^2 -15E-6/.15^2)
To solve this problem, we need to use the electric field formula, which states that the electric field at a point is equal to the force experienced by a positive test charge placed at that point divided by the magnitude of the test charge.
Step 1: Calculate the electric field due to each charge individually.
The electric field due to a point charge is given by the formula E = k(Q/r^2), where E is the electric field, k is the Coulomb's constant (9 × 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge.
For the left charge (Q1 = +25 μC) at 11 cm from the origin, the distance from the test point (15 cm to the right of the negative charge) is 26 cm. Convert this to meters: r1 = 26/100 = 0.26 m.
For the right charge (Q2 = -15 μC) at the origin, the distance from the test point is 15 cm. Convert this to meters: r2 = 0.15 m.
Step 2: Find the net electric field at the test point.
Since the electric field is a vector quantity, we need to consider both the magnitude and direction of the electric fields due to the two charges.
The electric field due to the positive charge will be towards the left because it is positive and repels positive charges.
The electric field due to the negative charge will be towards the right because it is negative and attracts positive charges.
To find the net electric field, we need to subtract the magnitude of the electric field due to the negative charge from the magnitude of the electric field due to the positive charge since they have opposite directions.
Step 3: Calculate the magnitudes of the individual electric fields.
Using the formula E = k(Q/r^2), we can calculate the electric field due to the left charge:
E1 = k(Q1/r1^2)
E1 = (9 × 10^9 Nm^2/C^2)(25 × 10^-6 C) / (0.26 m)^2
E1 ≈ 3.49 × 10^6 N/C
Now, let's calculate the electric field due to the right charge:
E2 = k(Q2/r2^2)
E2 = (9 × 10^9 Nm^2/C^2)(-15 × 10^-6 C) / (0.15 m)^2
E2 ≈ -6.0 × 10^6 N/C (negative because it is towards the right)
Step 4: Calculate the net electric field at the test point.
To find the net electric field, we subtract the magnitude of E2 from the magnitude of E1, but remember the directions matter too.
|E_net| = |E1| - |E2|
Since E1 is to the left and E2 is to the right, we add their magnitudes together:
|E_net| = |E1| + |E2|
|E_net| ≈ 3.49 × 10^6 N/C + 6.0 × 10^6 N/C
|E_net| ≈ 9.49 × 10^6 N/C
Therefore, the magnitude of the electric field at a point 15 cm to the right of the negative charge is approximately 9.49 × 10^6 N/C.