Magnitude and Direction of Electric Field

In summary, the problem involves two point charges (+3.0 μC and -2.0 μC) placed on the y-axis and a test charge (+1.0 x 10^-9 C) placed at different locations. The task is to find the magnitude and direction of the electric field and the force on the test charge at each location. Using the equation E = kq/(r^2), the force and electric field can be calculated for points (0, 1.0 m) and (0, -2.0 m) by considering the distance between the charges and the test charge. To find the force on the test charge at point (1.0 m, 0), the distance between (0
  • #1
bwilhelm
9
0

Homework Statement


Two point charges are placed on the y-axis. One is +3.0 C and located at position (0, 2 m), the other is -2.0 C and located at the origin. Use the diagram to find the magnitude and direction of the electric field and the force on a +1.0 x 10-9 C test charge placed in the following locations.
a. (0, 1.0 m)
b. (0, -2.0 m)
c. (1.0m, 0)

Homework Equations



E= kq/(r^2)

The Attempt at a Solution


So for part a:
F= (K(3x10^-6)/(1^2) + K(2x10-6)/(1^2))
E= F/(test charge)

Same for part b, but the radius is different.
My question is how do I get part C?
I calculated the force between the -2uc charge and point C, but how do I get the force between the +3uc charge and point C?

Do I need to find the angle?
 

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  • #2
Find the distance between points ( 0, 2) and (1,0).
Find field at (1,0) due to + 3μC and - 2μC.
Resolve the field due to +3μC into x and y components.
Find the net x component and find the resultant with y component.
 
  • #3
rl.bhat said:
Find the distance between points ( 0, 2) and (1,0).
Find field at (1,0) due to + 3μC and - 2μC.
Resolve the field due to +3μC into x and y components.
Find the net x component and find the resultant with y component.

Thank you for the reply, i'll try this out tomorrow.
 

Related to Magnitude and Direction of Electric Field

1. What is the definition of electric field magnitude and direction?

The magnitude of an electric field is the strength or intensity of the electric field at a particular point in space. It is a measure of the force experienced by a unit positive test charge placed at that point. The direction of the electric field is the direction in which the force would act on a positive test charge at that point.

2. How is the magnitude of an electric field calculated?

The magnitude of an electric field is calculated using the equation E = F/q, where E is the electric field, F is the force exerted on the test charge, and q is the magnitude of the test charge. The units for electric field are volts per meter (V/m).

3. What factors affect the magnitude of an electric field?

The magnitude of an electric field is affected by the distance from the source of the field, the magnitude of the source charge, and the type of material in between the source and the point where the electric field is measured. It is also influenced by the presence of other nearby charges.

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction in which a positive test charge would be pushed or pulled if placed at that point. The direction of the electric field is always perpendicular to the equipotential lines, which are imaginary lines that connect points with equal electric potential.

5. What are some real-life applications of understanding the magnitude and direction of electric field?

Understanding the magnitude and direction of electric fields is essential in many areas of science and technology. It is used in the design and operation of electronic devices, such as computers and cell phones. It is also crucial in fields such as electricity and magnetism, telecommunications, and power generation. Additionally, understanding electric fields is important in medical technology, such as in the use of MRI machines.

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