# The line integral of the electric field intensity gives the potential difference between two...

## Question:

The line integral of the electric field intensity gives the potential difference between two points: {eq}V_B - V_A = -\int _{P_A}^{P_B}\vec{E}(\vec{r})\cdot \vec{dl}. {/eq} If the electric field intensity is given in rectangular coordinates by {eq}\vec{E}(\vec{r})= 4y\hat{x} + 4x\hat{y} {/eq} V/m, find the potential difference between the points {eq}P_A(1, 1, 0) {/eq} and {eq}P_B (3, 3, 0) {/eq}.

Hint: Integrate along a straight line between the points and parameterize the path by x= y.

## Line Integrals:

Suppose that {eq}\vec{F} {/eq} is a vector field, and {eq}C {/eq} is an oriented curve contained in the domain of {eq}\vec{F} {/eq}. Let {eq}C {/eq} be parameterized by the differentiable function {eq}r(t) {/eq} where {eq}a \le t \le b {/eq}. Then the line integral of {eq}\vec{F} {/eq} over {eq}C {/eq} is defined to be

{eq}\displaystyle \int_C \vec{F} \cdot dr = \int_a^b \vec{F}(r(t)) \cdot r'(t) \, dt \, . {/eq}

The straight-line path {eq}C {/eq} from {eq}P_A {/eq} to {eq}P_B {/eq} can be parameterized by the function {eq}r(t)=(t,t,0) {/eq}, where {eq}1 \le t \le 3 {/eq}. Differentiating this parameterization gives {eq}r'(t)=\hat{x}+\hat{y} {/eq}. Assuming that coordinates are given in meters, the potential difference in volts is therefore:

\begin{align*} -\iint_C \vec{E} \cdot dr &= -\int_1^3 \vec{E}(r(t)) \cdot r'(t) \, dt&&\text{(by the definition of the line integral)}\\ &=-\int_1^3 \vec{E}(t, t, 0) \cdot (\hat{x} + \hat{y}) \, dt&&\text{(by the above expressions for }x,y\text{)}\\ &=-\int_1^3 (4t\hat{x}+4t\hat{y}) \cdot (\hat{x}+\hat{y}) \, dt&&\text{(by the given expression for }\vec{E}\text{)}\\ &=-\int_1^3 \left[4t(1)+4t(1)\right] \, dt&&\text{(evaluating the dot product)}\\ &=-\int_1^3 8t \, dt\\ &=-4t^2\Big|_1^3&&\text{(evaluating the integral)}\\ &=-4(3^2)+4(1^2)\\ &=-32 \, . \end{align*}

So the potential difference is {eq}\boxed{-32\text{ V}}\, {/eq}.