Electrostatics Part 3

Concept:Electric potential difference between two points is the negative of the work done by the electric field in moving a unit charge from one point to the other.Explanation:The potential difference is given by: $V_2 - V_1 = -\int \vec{E} \cdot d\vec{r}$.Here, $\vec{E} = A x \hat{i} + B y \hat{j}$ with $A = 10 \, \mathrm{V/m^2}$, $B = 5 \, \mathrm{V/m^2}$.Point 1 is the origin $(0,0)$ with unknown potential $V_0$.Point 2 is $(10,20)$ with potential $500 \, \mathrm{V}$.$d\vec{r} = dx \hat{i} + dy \hat{j}$.So, $V_2 - V_0 = -\int_{(0,0)}^{(10,20)} (A x\, dx + B y\, dy)$.Separate integrals: $V_2 - V_0 = - \left[ \int_0^{10} A x\, dx + \int_0^{20} B y\, dy \right]$.Calculate: $\int_0^{10} A x\, dx = A \cdot \frac{10^2}{2} = 10 \times 50 = 500$.Similarly: $\int_0^{20} B y\, dy = B \cdot \frac{20^2}{2} = 5 \times 200 = 1000$.Thus: $500 - V_0 = - (500 + 1000) = -1500$.Solve: $V_0 = 500 + 1500 = 2000 \, \mathrm{V}$.Answer:$2000 \, \mathrm{V}$, which corresponds to option C.