Question

007 10.0 points An electric field is given by Ex = (5 kN/C) x 3 . Find the potential difference between the points on the x-axis at x = 1 m and x = 2 m. Answer in units of kV.

Answer 1

Here

$E_{x}=5x^{3}kN/C$

We know

Potential difference $=\Delta V$

$\Delta V=-\int E.dx$

$\Delta V=-\int_{x=1}^{x=2}E.dx=-\int_{x=1}^{x=2}5x^{3}.dx=-\frac{5}{4}(2^{4}-1^{4})kV=-18.75kV$