An electric field $$\vec{E} = \left(25\hat{i} + 30\hat{j}\right)$$ N C$$^{-1}$$ exists in a region of space. If the potential at the origin is taken to be zero then the potential at $$x = 2$$ m, $$y = 2$$ m is:

We have been given a uniform electric field $$\vec{E} = 25\hat{i} + 30\hat{j}\;{\rm N\,C^{-1}}$$. The potential at the origin $$O(0,0)$$ is specified to be zero, that is $$V_O = 0$$.

For any two points $$A$$ and $$B$$ in an electrostatic field, the potential difference is obtained from the basic relation

$$V_B - V_A = -\int_{A}^{B} \vec{E}\cdot d\vec{l}.$$

Here, point $$A$$ is the origin $$O(0,0)$$ and point $$B$$ is $$P(2\,{\rm m},\,2\,{\rm m})$$. Because the field is uniform (its components do not depend on position), we can choose any convenient path; the line integral of a constant vector reduces to the dot-product of that vector with the overall displacement. Hence,

$$V_P - V_O = -\vec{E}\cdot\vec{r},$$

where $$\vec{r}$$ is the displacement vector from $$O$$ to $$P$$.

The displacement vector is obtained component-wise:

$$\vec{r} = (2 - 0)\hat{i} + (2 - 0)\hat{j} = 2\hat{i} + 2\hat{j}\;{\rm m}.$$

Next we compute the dot product $$\vec{E}\cdot\vec{r}$$ step by step:

$$\vec{E}\cdot\vec{r} = (25\hat{i} + 30\hat{j}) \cdot (2\hat{i} + 2\hat{j})$$

$$\;\;\;= 25 \times 2 + 30 \times 2$$

$$\;\;\;= 50 + 60$$

$$\;\;\;= 110.$$

Now we insert this result into the potential difference relation:

$$V_P - V_O = -(\vec{E}\cdot\vec{r}) = -110.$$

But $$V_O = 0$$ by the condition given, so

$$V_P = -110 \;{\rm J\,C^{-1}}.$$

Hence, the correct answer is Option A.