If `V=2r^(2)` then find (i) `vec(E) (1, 0, -2)` (ii) `vec(E) (r=2)`

Answer

To solve the given problem, we need to find the electric field vector \(\vec{E}\) based on the potential \(V = 2r^2\). We will tackle the problem in two parts as specified.

### Step-by-Step Solution

**Part (i): Find \(\vec{E}\) at the point (1, 0, -2)**

1. **Understand the relationship between electric field and potential**:
The electric field \(\vec{E}\) is related to the electric potential \(V\) by the equation:
\[ \vec{E} = -\nabla V \]
where \(\nabla V\) is the gradient of the potential.

2. **Calculate the gradient of \(V\)**:
Given \(V = 2r^2\), where \(r^2 = x^2 + y^2 + z^2\), we can express \(V\) in terms of \(x\), \(y\), and \(z\):
\[ V = 2(x^2 + y^2 + z^2) \]
Now, we calculate the partial derivatives:
\[ \frac{\partial V}{\partial x} = 4x, \quad \frac{\partial V}{\partial y} = 4y, \quad \frac{\partial V}{\partial z} = 4z \]

3. **Write the electric field vector**:
Thus, the electric field vector \(\vec{E}\) can be expressed as:
\[ \vec{E} = -\left(4x \hat{i} + 4y \hat{j} + 4z \hat{k}\right) = -4x \hat{i} - 4y \hat{j} - 4z \hat{k} \]

4. **Evaluate \(\vec{E}\) at the point (1, 0, -2)**:
Substitute \(x = 1\), \(y = 0\), and \(z = -2\):
\[ \vec{E} = -4(1) \hat{i} - 4(0) \hat{j} - 4(-2) \hat{k} = -4 \hat{i} + 8 \hat{k} \]
Therefore, the electric field at the point (1, 0, -2) is:
\[ \vec{E} = -4 \hat{i} + 8 \hat{k} \]

**Part (ii): Find \(\vec{E}\) at \(r = 2\)**

1. **Use the relationship between \(V\) and \(r\)**:
We know \(V = 2r^2\). To find the electric field in terms of \(r\), we differentiate \(V\) with respect to \(r\):
\[ \frac{dV}{dr} = \frac{d}{dr}(2r^2) = 4r \]

2. **Find \(\vec{E}\)**:
The electric field in terms of \(r\) is:
\[ \vec{E} = -\frac{dV}{dr} = -4r \]

3. **Evaluate \(\vec{E}\) at \(r = 2\)**:
Substitute \(r = 2\):
\[ \vec{E} = -4(2) = -8 \]
Thus, the magnitude of the electric field at \(r = 2\) is:
\[ \vec{E} = -8 \]

### Final Answers
(i) \(\vec{E} = -4 \hat{i} + 8 \hat{k}\) at the point (1, 0, -2)
(ii) \(\vec{E} = -8\) at \(r = 2\)