An electric field is given by Ex = (2 kN/C) x^3.?

Question

Find the potential difference between the points on the x-axis at x = 1 m and x = 2 m. Answer in units of kV.

To begin this problem, I simply found the electric field ateach point by plugging in the two distances into the E-field equation. After this, I really wasn't sure where to go. I know that potential difference is equal to the integral of the dot product between the E-field and distance. Unfortunately, I'm not sure how to apply this concept to this problem. Maybe integrating Ex would help? Any help would be appreciated.

Answer 1

OK, The idea es practically the same, you have to remember, that the potential difference, is : - integration of (E.dr)

So, here you have : E = 2 k N/C * x^3

And x goes from 1 to 2

Potential difference, will be : - integration(2*x^3)

when you integrate it, you will have : 2*x^4 / 4 = x^4 / 2

now, using the values of x, from 1 to 2

2^4 / 2 - 1^4 */ 2 = 8 - 1/2 = 15 / 2

with the minus : - 15 / 2

the potential differences : -15 / 2 KVolts

Answer 2

This question cannot be answered in units of kV,
simply because the coefficient has improper dimension.

The correct dimension of η in Ex = ηx^3 must be N/Cm³.
There is no way to convert 2 kN/C into N/Cm³.

Answer 3

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