The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero

Options

A. It is uniform in the region
B. It is proportional to r
C. It is proportional to `r^(2)`
D. It increase as one goes away from the origin.

Answer

To solve the problem, we need to find the electric potential \( V \) at a distance \( r \) from the origin, given that the electric field \( E \) is directed outward and is proportional to the distance \( r \) from the origin. We will also use the fact that the electric potential at the origin is zero.

### Step-by-Step Solution:

1. **Understanding the Electric Field**:
We know that the electric field \( E \) is given to be proportional to the distance \( r \) from the origin. We can express this mathematically as:
\[ E = k \cdot r \]
where \( k \) is a constant of proportionality.

2. **Relating Electric Field to Electric Potential**:
The relationship between electric field \( E \) and electric potential \( V \) is given by:
\[ E = -\frac{dV}{dr} \]
This means that the electric field is the negative gradient of the electric potential.

3. **Setting Up the Integral**:
Since \( E \) is directed outward, we can write:
\[ -\frac{dV}{dr} = k \cdot r \]
Rearranging gives:
\[ dV = -k \cdot r \, dr \]

4. **Integrating to Find Potential**:
To find the potential \( V \) at a distance \( r \) from the origin, we need to integrate:
\[ V = \int_0^r -k \cdot r' \, dr' \]
Here, \( r' \) is a dummy variable for integration.

5. **Calculating the Integral**:
The integral becomes:
\[ V = -k \int_0^r r' \, dr' = -k \left[ \frac{(r')^2}{2} \right]_0^r = -k \cdot \frac{r^2}{2} \]
Thus, we have:
\[ V = -\frac{k}{2} r^2 \]

6. **Final Expression for Electric Potential**:
Since we are given that the electric potential at the origin is zero, we can conclude that:
\[ V(r) = -\frac{k}{2} r^2 \]

### Summary:
The electric potential \( V \) at a distance \( r \) from the origin is given by:
\[ V(r) = -\frac{k}{2} r^2 \]