How Does the Electric Field Influence Potential Difference Calculation?

[Arman777]

Homework Statement

$V = V (x,y,z) = 10^3(2x^2 + 3y^2 −5z^2 + 7x)−19999$
Find $V_{ba}$\vec {r_a}=2i##
$\vec {r_b}=j+k$

Homework Equations

$E$ will be $E=-∇V$
$V_b-V_a=-\int_a^b E(r)\, dr$

The Attempt at a Solution

When I try to calculate the potential using $V_b-V_a=-\int_a^b E(r)dr$, and $E=-∇V$ dr I get $-26.10^3$
Electric field at a is $-15.10^3i$
Electrci Field at b is $-6.10^3j+10.10^3k$
so inital $V$ is $-30.10^3 V$ final $V$ is $-4.10^3 V$ from integral and thres minus sign so its $-26.10^3 V$
when I use $V = V (x,y,z) = 10^3(2x^2 + 3y^2 −5z^2 + 7x)−19999$ inserting here I get $-24.10^3V$

 

[TSny]

$V_b-V_a=-\int_a^b E(r)dr$

The general expression is $V_b-V_a=-\int_a^b \vec{E} \cdot d\vec{s}$, where the integration is along some path connecting the points a and b. The integrand involves a dot product.

Electric field at a is $-15.10^3i$
Electrci Field at b is $-6.10^3j+10.10^3k$
so inital $V$ is $-30.10^3 V$ final $V$ is $-4.10^3 V$ from integral and thres minus sign so its $-26.10^3 V$

It's not clear how you did the integration. What path did you choose? Why did you evaluate $\vec E$ at the specific points a and b? The integral depends on $\vec E$ at all points of the path of integration.