Electric Field As Derivative of Potential Video Lessons

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Problem: The electric potential in a region of space is V=350V?mx2+y2?, where x and y are in meters.Part AWhat is the strength of the electric field at (x,y) = (2.3m, 2.9m)?Express your answer using two significant figures.Part BWhat is the direction of the electric field at (x,y) = (2.3m, 2.9m)? Give the direction as an angle ccw from the positive x-axis.Express your answer using two significant figures.

FREE Expert Solution

The electric field is given by:

E=-V=-(xi^+yj^)V

Vector Magnitude:

|A|=Ax2+Ay2

Direction:

tanθ=AyAx

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Problem Details

The electric potential in a region of space is V=350V?mx2+y2?, where x and y are in meters.



Part A

What is the strength of the electric field at (x,y) = (2.3m, 2.9m)?

Express your answer using two significant figures.


Part B

What is the direction of the electric field at (x,y) = (2.3m, 2.9m)? Give the direction as an angle ccw from the positive x-axis.

Express your answer using two significant figures.

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