If the electric potential along the x-axis is V = 500 x2𝑥2 V, where x is in meters, what is Ex𝐸𝑥 at x = 0.500 m? 250. V/m -250. V/m -500. V/m 500. V/m
Question
If the electric potential along the x-axis is V = 500 x2𝑥2 V, where x is in meters, what is Ex𝐸𝑥 at x = 0.500 m? 250. V/m -250. V/m -500. V/m 500. V/m
Solution
The electric field E is related to the potential V by the equation E = -dV/dx, where dV/dx is the derivative of the potential with respect to position.
Given V = 500x^2, we can differentiate this with respect to x to find the electric field.
dV/dx = 2500x = 1000x
Substituting x = 0.500 m into this equation gives:
E = -dV/dx = -1000*0.500 = -500 V/m
So, the electric field Ex at x = 0.500 m is -500 V/m.
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