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What is ex(p), the value of the x-component of the electric field produced by q1 and q2 at point p??

2 Answers

4 votes

Final answer:

The x-component of the electric field at point P is determined by calculating the vector sum of the fields due to each charge. The force on a particle in an electric field is given by the product of the charge and the electric field strength. For uniform fields, the potential difference is calculated by the product of the charge, electric field strength, and distance moved in the field direction.

Step-by-step explanation:

To calculate the x-component of the electric field (Ex) produced by charges q1 and q2 at point P, we use Coulomb's Law. For a point charge, the electric field E at a distance r from the charge is given by:

E = k * |q| / r2

where k is Coulomb's constant (8.99x109 Nm2/C2), q is the charge, and r is the distance from the charge to the point of interest. Since the electric field is a vector quantity, its components Ex, Ey, and Ez can be calculated by decomposing E into its respective components. For two charges, q1 and q2, the net electric field at point P is the vector sum of the electric fields due to each charge individually. This takes into account both the magnitude and direction (angle).

The force F exerted by an electric field E on a charge q is given by:

F = qE

So, if we are considering the force on a +2q charged particle due to the electric field produced by a +q charge, it would be twice the electric field strength, assuming the distance from the charge to the point P remains the same.

In regards to uniform electric fields, they do not change with position (x), and thus the x-component Ex would be constant. This is different from nonuniform fields, which change with position and need to be calculated considering the specific geometry of the setup.

An electric potential difference (ΔU) in a uniform electric field can also be calculated using the following relationship, showing the work done in moving a charge q through a distance in the direction of the electric field:

ΔU = −qE(xf − xi)

Here, xf and xi are the final and initial positions, respectively.

User Steakchaser
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8.5k points
4 votes
you will have Ex(P)=(9E9)*(-3.9E-6)/(.067^2+.067^2).
this will give you the magnitude of the electric field produced by q1. all thats left is to multiply your answer by a sin(45) and voila, there's your answer!
This should be your final answer (-2764474.05 N/C)

for the second part you do the same, use the formula but this time q2 will affect it because you are looking for the y-component.

Ey(P)=(9E9)*(7.1E-6)/(.067^2) which equals out to 14234796.17 N/C.
to get the final answer just use the answer from part A and subtract it from the above answer and you are all set to go!
This should be the final answer for the second part. (11470322.12 N/C)
User BigGinDaHouse
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7.6k points