Finding the components of an electric field.

In summary, to find the electric field at a point where a 1.05μC charge experiences a force of (3.0i-3.9j)x10^-3 N, we need to first solve for the vector sum of the force components. Then, using the equation E=F/q, we can find the electric field by dividing the force vector by the charge scalar. This can also be done component-by-component to get the individual components of the electric field vector.
  • #1
bfusco
128
1

Homework Statement


What is the electric field at a point when the force on a 1.05μC charge placed at that point is (3.0i-3.9j)x10^-3 N? (E=αi+βj N/C).
a) Find α
b) Find β

The Attempt at a Solution


first i took the force and solved for the vector sum of the components:
√[(3x10^-3)^2 + (3.9x10^-3)^2]→ 4.92x10-3 so F=4.92x10^-3

to find the electric field the equation is E=F/q:
E=(4.92x10^-3)/(1.05x10^-6)=4685, assuming that is all correct, how do i get the components of an electric field? basically with 4685 how do i get the components?
 
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  • #2
Isn't the electric field intensity vector equal to the vector force divided by the charge?
 
  • #3
yes...E=F/q, which i used
 
  • #4
Yes, but this equation also applies component-by-component. F is a vector and q is a scalar. F divided by q, component-by-component, gives the new vector E. This is how you get the individual components.
 
  • #5


I would suggest using the equation E=F/q to find the magnitude of the electric field, which you have correctly calculated as 4685 N/C. To find the components of the electric field, you can use the equation E=αi+βj N/C, where α and β represent the components in the x and y directions respectively.

To find α, you can use the fact that the x-component of the electric field is equal to the x-component of the force divided by the charge (α=F_x/q). Similarly, you can find β by dividing the y-component of the force by the charge (β=F_y/q).

In this case, we already have the x and y components of the force (3.0i-3.9j)x10^-3 N, so we can simply divide each component by the charge of 1.05μC to find the corresponding component of the electric field. Thus, α=2.86x10^3 N/C and β=-3.71x10^3 N/C.

In summary, the components of the electric field at the given point are α=2.86x10^3 N/C in the x-direction and β=-3.71x10^3 N/C in the y-direction.
 

1. What is an electric field and why is it important?

An electric field is a physical quantity that describes the effect of electric forces on charged particles. It is important because it helps us understand and predict the behavior of electrically charged objects and plays a crucial role in many technological applications.

2. How do you find the components of an electric field?

The components of an electric field can be found by using mathematical equations, such as Coulomb's law or Gauss's law, which take into account the magnitude and direction of the electric field at a given point.

3. What factors affect the strength and direction of an electric field?

The strength and direction of an electric field can be affected by the magnitude and distribution of the electric charges, as well as the distance between the charges. The presence of conductors or insulators can also influence the electric field.

4. How is an electric field different from an electric potential?

While an electric field is a vector quantity that describes the force on a charged particle, the electric potential is a scalar quantity that describes the potential energy per unit charge at a given point in the electric field. In simpler terms, the electric field tells us which direction a charged particle will move, while the electric potential tells us how much work is required to move the charged particle.

5. What are some real-life examples of electric fields?

Electric fields are present in many everyday objects and technologies, such as batteries, power lines, and electronic devices. They are also essential in biological systems, as they play a role in nerve impulses and muscle contractions.

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