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Sagot :
The magnitude of the uniform electric field, if the electric potential is -2000 v at x = 8m and is 400 v at x = 2 m in x direction will be, 400V/m.
To find the answer, we have to know more about the electric field.
How to find the magnitude of uniform electric field?
- a space where other charged particles or things would be forced to move away from a charged particle or item is called the electric field.
- The expression for electric field in terms of potential V and the distance x can be written as,
[tex]E=\frac{dV}{dx}[/tex]
where, dV is the potential difference at two different points of the conductor and dx is the difference of the points.
- Thus, the magnitude of electric field can be written as,
[tex]|E|=|-(\frac{dV}{dx} )|=|-(\frac{-2000-400}{8-2} )|\\\\|E|=400V/m[/tex]
Thus, we can conclude that, the magnitude of the uniform electric field, if the electric potential is -2000 v at x = 8m and is 400 v at x = 2 m in x direction will be, 400V/m.
Learn more about the electric field here:
https://brainly.com/question/27752270
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