Concept:
The electric field is the negative gradient of the potential: \(\vec{E} = -\nabla V\). Therefore, the potential difference can be found by integrating the field: \(V = -\int \vec{E} \cdot d\vec{r}\).
Step 1: Integrate to find the potential function \(V(x,y)\).
\[ dV = -E_x dx - E_y dy = -(10x)dx - (5y)dy \]
\[ V(x,y) = \int -10x dx - \int 5y dy \]
\[ V(x,y) = -5x^2 - \frac{5}{2}y^2 + C \]
Step 2: Determine constant C.
We are given \(V(10, 20) = 500\).
\[ 500 = -5(10)^2 - \frac{5}{2}(20)^2 + C \]
\[ 500 = -500 - \frac{5}{2}(400) + C \]
\[ 500 = -500 - 1000 + C \]
\[ 500 = -1500 + C \implies C = 2000 \]
Step 3: Calculate potential at origin.
At \((0,0)\):
\[ V(0,0) = -5(0) - \frac{5}{2}(0) + C = 2000\,\text{V} \]
\[
\boxed{V_{\text{origin}} = 2000\ \text{volt}}
\]