An electric field `E = (20hati + 30 hatj)` N/C exists in the space. If the potential at the origin is taken be zero, find the potential at `(2 m, 2 m)`.

Explanation

To find the electric potential at the point (2 m, 2 m) given the electric field \( \mathbf{E} = (20 \hat{i} + 30 \hat{j}) \) N/C and the potential at the origin being zero, we can follow these steps:

### Step 1: Understand the relationship between electric field and potential
The change in electric potential \( V \) between two points in an electric field is given by the formula:
\[ V = V_0 - \int_{\mathbf{r_0}}^{\mathbf{r}} \mathbf{E} \cdot d\mathbf{r} \]
where \( V_0 \) is the potential at the initial point, \( \mathbf{E} \) is the electric field, and \( d\mathbf{r} \) is the differential displacement vector.

### Step 2: Identify the points
In this case, we have:
- The initial point (origin) \( \mathbf{r_0} = (0, 0) \) with \( V_0 = 0 \).
- The final point \( \mathbf{r} = (2, 2) \).

### Step 3: Calculate the displacement vector
The displacement vector \( d\mathbf{r} \) from the origin to the point (2 m, 2 m) is:
\[ d\mathbf{r} = (2 - 0) \hat{i} + (2 - 0) \hat{j} = 2 \hat{i} + 2 \hat{j} \]

### Step 4: Calculate the dot product
Now, we need to calculate the dot product \( \mathbf{E} \cdot d\mathbf{r} \):
\[ \mathbf{E} = 20 \hat{i} + 30 \hat{j} \]
\[ d\mathbf{r} = 2 \hat{i} + 2 \hat{j} \]
The dot product is:
\[ \mathbf{E} \cdot d\mathbf{r} = (20 \hat{i} + 30 \hat{j}) \cdot (2 \hat{i} + 2 \hat{j}) = 20 \cdot 2 + 30 \cdot 2 = 40 + 60 = 100 \]

### Step 5: Calculate the potential at (2 m, 2 m)
Now, substituting the dot product into the potential equation:
\[ V = V_0 - \mathbf{E} \cdot d\mathbf{r} \]
\[ V = 0 - 100 = -100 \text{ volts} \]

### Conclusion
The potential at the point (2 m, 2 m) is:
\[ \boxed{-100 \text{ volts}} \]