What are are the magnitude and direction

[gbedenba]

Homework Statement

The quantity called the electric field is a vector. The electric field inside a scientific instrument is = (215 - 380 ) V/m, where V/m stands for volts per meter. What are the magnitude and direction of the electric field?

_____ v/m

______ degree (smallest positive angle from the positive x-axis)

Homework Equations

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The Attempt at a Solution

I don't know where to start

 

[tiny-tim]

Welcome to PF!

Hi gbedenba! Welcome to PF!

The quantity called the electric field is a vector. The electric field inside a scientific instrument is = (215 - 380 ) V/m, where V/m stands for volts per meter. What are the magnitude and direction of the electric field?

Do you mean that the x and y coordinates are 215 and -380?

Use Pythagoras, and tangent.

 

[tomcue]

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I would first clarify the context of the problem. Is this electric field measured inside the instrument or at a specific point inside the instrument? Is there any information about the geometry or orientation of the instrument? This information is necessary to accurately determine the magnitude and direction of the electric field.

Assuming that the electric field is being measured at a specific point inside the instrument, we can use the given vector components (215 V/m and -380 V/m) to calculate the magnitude and direction of the electric field. The magnitude can be calculated using the Pythagorean theorem:

|E| = √(215^2 + (-380)^2) = 441.67 V/m

The direction can be determined by finding the angle between the electric field vector and the positive x-axis:

θ = tan^-1 (-380/215) = -60.38 degrees

However, since the angle is measured counterclockwise from the positive x-axis, the smallest positive angle would be 360 - 60.38 = 299.62 degrees.

Therefore, the magnitude of the electric field is 441.67 V/m and the direction is 299.62 degrees counterclockwise from the positive x-axis.