Electric potential at origin is zero. Field in that space is `10hati+10hatjNC^(-1)`. Find electric potential at (1, 1).

Answer

To find the electric potential at the point (1, 1) given that the electric potential at the origin is zero and the electric field in that space is \( \mathbf{E} = 10 \hat{i} + 10 \hat{j} \, \text{N/C} \), we can follow these steps:

### Step 1: Understand the relationship between electric potential and electric field
The electric potential \( V \) at a point in an electric field can be calculated using the formula:
\[ V = -\int \mathbf{E} \cdot d\mathbf{r} \]
where \( d\mathbf{r} \) is the differential displacement vector.

### Step 2: Define the displacement vector
In this case, we will consider a path from the origin (0, 0) to the point (1, 1). The displacement vector can be expressed as:
\[ d\mathbf{r} = dx \hat{i} + dy \hat{j} \]

### Step 3: Substitute the electric field and displacement vector into the integral
Given \( \mathbf{E} = 10 \hat{i} + 10 \hat{j} \), we can calculate the dot product:
\[ \mathbf{E} \cdot d\mathbf{r} = (10 \hat{i} + 10 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 10 dx + 10 dy \]

### Step 4: Set up the integral for electric potential
Now we can set up the integral from the origin (0, 0) to (1, 1):
\[ V = -\int_{(0,0)}^{(1,1)} (10 dx + 10 dy) \]

### Step 5: Evaluate the integral
We can break this integral into two parts:
\[ V = -\left( \int_{0}^{1} 10 \, dx + \int_{0}^{1} 10 \, dy \right) \]
Calculating each integral:
1. For \( dx \):
\[ \int_{0}^{1} 10 \, dx = 10 \cdot [x]_{0}^{1} = 10(1 - 0) = 10 \]
2. For \( dy \):
\[ \int_{0}^{1} 10 \, dy = 10 \cdot [y]_{0}^{1} = 10(1 - 0) = 10 \]

Combining these results:
\[ V = -\left(10 + 10\right) = -20 \, \text{V} \]

### Final Answer
Thus, the electric potential at the point (1, 1) is:
\[ V(1, 1) = -20 \, \text{V} \]

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