Electric Field As Derivative of Potential Video Lessons

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Problem: A given system has the equipotential surfaces shown in the figure (Figure 1).a) What is the magnitude of the electric field? (V/m)b) What is the direction of the electric field? ( ∘ from the +x axis )

FREE Expert Solution

Electric field components:

Ex=-VxEy=-Vy

The magnitude of the electric field:

E=Ex2+Ey2

Direction:

θ=tan-1(EyEx)

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Problem Details

A given system has the equipotential surfaces shown in the figure (Figure 1).

a) What is the magnitude of the electric field? (V/m)

b) What is the direction of the electric field? ( ∘ from the +x axis )


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