Magnitude and direction of Electric field

In summary, the magnitude of the electric field is 17.8 N/C and the direction is 308.2 degrees counterclockwise from the +x-axis, which is found by taking the inverse tangent of 14/-11 and subtracting it from 180 degrees.
  • #1
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Homework Statement



Calculate the magnitude of the electric field E = (-11i^+14j^)N/C


∣∣E⃗ ∣∣ = _______ N/C


Calculate the direction (relative to the +x-axis ) of the electric field E = (-11i^+14j^)N/C.

θ =_________∘ counterclockwise from the +x-axis

Homework Equations





The Attempt at a Solution



So magnitude would be SQRT(-11^2 + 14^2) = SQRT(317) = 17.8 N/C

For direction you'd take the inverse tan so:

theta = ATAN(14/-11) = -51.8 degrees

But that means the angle is 51.8 degrees BELOW the x-axis, so counter-clockwise it'd be

360-51.8 = 308.2 degrees
BUT I'm getting this part wrong and I don't know why? what am I doing wrong
 
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  • #2
From the expression for the field vector in component form, it looks like the angle lies in the second quadrant.
 
  • #3
So should i just subtract 51.8 from 180 degrees?
 
  • #4
vanitymdl said:
So should i just subtract 51.8 from 180 degrees?
Yes.
 
  • #5
Thank you!
 
  • #6
How do you know to find theta using arctan(14/11)? and then to subtract that from 180?
 

What is the magnitude of an electric field?

The magnitude of an electric field is a measure of the strength of the force exerted by the electric field on a charged particle at a specific point in space. It is typically represented by the symbol E and is measured in units of Newtons per Coulomb (N/C).

How is the magnitude of an electric field calculated?

The magnitude of an electric field can be calculated by dividing the force exerted on a test charge by the magnitude of the charge itself. This can be written as: E = F/q, where E is the electric field, F is the force, and q is the test charge. Alternatively, it can also be calculated by dividing the electric potential difference between two points by the distance between those points.

What factors affect the magnitude of an electric field?

There are several factors that can affect the magnitude of an electric field. These include the amount of charge present, the distance between the charges, and the medium through which the electric field is passing. Additionally, the direction of the electric field and the presence of other charged particles can also impact its magnitude.

What is the direction of an electric field?

The direction of an electric field is determined by the direction in which a positive test charge would move if placed in the electric field. It is typically represented by arrows, with the length of the arrow indicating the strength of the field. The direction of the electric field is always perpendicular to the equipotential lines and points towards the negative charge and away from the positive charge.

How is the direction of an electric field determined?

The direction of an electric field can be determined by the principle of superposition, which states that the total electric field at a point is equal to the vector sum of the individual electric fields caused by each individual charge. This means that the direction of the electric field can be determined by analyzing the direction and magnitude of each individual charge in the field.

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