Calculating Potential Difference in an Electric Field with Given Field Equation

AI Thread Summary
The electric field is defined by the equation Ex = 2.2x + 5x² kN/C, and the task is to calculate the potential difference between points at x = -2.0 m and x = 2.0 m. The integral used for the calculation is ∫(5x² + 2.2x) from -2 to 2. The initial calculation yielded 27V, but the correct answer is -27V due to the negative sign in the relationship between electric field and potential difference, as expressed by the formula ΔV = -∫E·ds. The oversight was in neglecting the negative sign in the integral, which is crucial for accurate results.
chopnhack
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Homework Statement


An electric field is given by Ex = 2.2x + 5x2kNC-1. Find the potential difference between two points on the x-axis at x = -2.0 m and x = 2.0 m

Homework Equations


∫5x2+2.2x from -2 to 2

The Attempt at a Solution


I got the answer of 27V.

Solution said - negative 27V

Can someone please explain?
Thanks
 
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chopnhack said:
∫5x2+2.2x from -2 to 2
Remember there is a negative sign in front of this integral since
$$\mathbf{E}=-\nabla V$$.
 
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##\Delta V = -\int_{-2}^2{\vec{E} \cdot d\vec{s}}##
You left out the negative sign.
 
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NFuller said:
Remember there is a negative sign in front of this integral since
$$\mathbf{E}=-\nabla V$$.
yes, sadly I recall it now...

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I was so happy to have found the proper way of handling it, I didn't recall the formula!
Thanks all.
 
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