have homework due soon, need help thanks.
Two point charges are placed on the x axis.(Figure 1) The first charge, q1 = 8.00nC , is placed a distance 16.0m from the origin along the positive x axis; the second charge, q2 = 6.00nC , is placed a distance 9.00m from the origin along the negative x axis.
Part A
Find the electric field at the origin, point O.
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.
Part B
Now, assume that charge q2 is negative; q2=−6nC. (Figure 2) What is the net electric field at the origin, point O?
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.
To find the electric field at the origin (point O) in both Part A and Part B of this question, we need to use the principles of electrostatics and the formula for electric field due to a point charge.
The formula to find the electric field due to a point charge is given by:
E = k * q / r^2
- E is the magnitude of the electric field
- k is the electrostatic constant, approximately equal to 8.99 x 10^9 N m^2/C^2
- q is the magnitude of the point charge
- r is the distance from the point charge to the point where the electric field is being measured
Part A:
In this part, both charges are positive, i.e., q1 = 8.00 nC and q2 = 6.00 nC.
The electric field at the origin, caused by q1, can be calculated using the formula as follows:
E1 = k * q1 / r1^2
Here, q1 = 8.00 nC and r1 = 16.0 m. Plug in these values to get E1.
Similarly, the electric field at the origin, caused by q2, can be calculated using the formula as follows:
E2 = k * q2 / r2^2
Here, q2 = 6.00 nC and r2 = 9.00 m. Plug in these values to get E2.
Now, since the electric fields due to both point charges act along the x-axis, they can be added together as vectors. The resultant electric field at the origin is given by the sum of E1 and E2.
The x-component of the electric field is given by the sum of the x-components of E1 and E2. The y-component of the electric field is zero since the charges are placed along the x-axis.
Part B:
In this part, the charge q2 is negative, i.e., q2 = -6.00 nC.
Repeat the calculations for E1 and E2, as in Part A, but with q2 equal to -6.00 nC.
Find the resultant electric field at the origin, same as in Part A, by summing the x-components of E1 and E2. The y-component remains zero.
By following these steps and calculating the appropriate values, you'll be able to find the answers to both Part A and Part B of the question.