Electric field in a region is given by $$\overrightarrow{E}=Ax\widehat{i}+By\widehat {j}$$, where $$A= 10V/m^{2}$$, and $$B= 5V/m^{2}$$. If the electric potential at a point (10, 20) is 500 $$V$$, then the electric potential at origin is____ $$V$$.

We use the relation between electric field and potential: $$\overrightarrow{E} = -\nabla V\,. $$ For a two-dimensional field $$\overrightarrow{E}=E_x\,\widehat{i}+E_y\,\widehat{j}$$, this gives

$$E_x = -\frac{\partial V}{\partial x},\quad E_y = -\frac{\partial V}{\partial y}\,.$$

Given $$E_x = A\,x$$ and $$E_y = B\,y$$, we write:

$$-\frac{\partial V}{\partial x} = A\,x\quad\Longrightarrow\quad \frac{\partial V}{\partial x} = -A\,x \quad -(1)$$

Integrate $$(1)$$ with respect to $$x$$ (treating $$y$$ as constant):

$$V(x,y) = -A\int x\,dx + f(y) = -\frac{A\,x^2}{2} + f(y)\quad -(2)$$

Similarly, from $$E_y$$:

$$-\frac{\partial V}{\partial y} = B\,y\quad\Longrightarrow\quad \frac{\partial V}{\partial y} = -B\,y \quad -(3)$$

Differentiate expression $$(2)$$ with respect to $$y$$:

$$\frac{\partial V}{\partial y} = \frac{d}{dy}\Bigl(-\frac{A\,x^2}{2} + f(y)\Bigr) = f'(y)\quad -(4)$$

Equate $$(3)$$ and $$(4)$$:

$$f'(y) = -B\,y$$

Integrate with respect to $$y$$:

$$f(y) = -B\int y\,dy + C = -\frac{B\,y^2}{2} + C\quad -(5)$$

Substitute $$(5)$$ into $$(2)$$ to obtain the general potential:

$$V(x,y) = -\frac{A\,x^2}{2} - \frac{B\,y^2}{2} + C\quad -(6)$$

Use the condition $$V(10,20)=500\,$$V in $$(6)$$:

$$500 = -\frac{A\,(10)^2}{2} - \frac{B\,(20)^2}{2} + C\,. $$

Substitute $$A=10\,$$V/m2 and $$B=5\,$$V/m2:

$$500 = -\frac{10\times100}{2} - \frac{5\times400}{2} + C = -500 -1000 + C\,. $$

Thus,

$$C = 500 + 1500 = 2000\quad -(7)$$

The potential at the origin $$(0,0)$$ is:

$$V(0,0) = -\frac{A\cdot0^2}{2} - \frac{B\cdot0^2}{2} + C = C = 2000\;\text{V}\,.$$

Answer: 2000