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 \(\vec{E}\) at two different points based on the potential \(V = 2r^2\).

### Step 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.

### Step 2: Calculate the gradient of the potential
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 compute the gradient \(\nabla V\):
\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]

Calculating each component:
- \(\frac{\partial V}{\partial x} = 4x\)
- \(\frac{\partial V}{\partial y} = 4y\)
- \(\frac{\partial V}{\partial z} = 4z\)

Thus, we have:
\[ \nabla V = (4x, 4y, 4z) \]

### Step 3: Find the electric field \(\vec{E}\)
Now substituting into the equation for \(\vec{E}\):
\[ \vec{E} = -\nabla V = (-4x, -4y, -4z) \]

### Step 4: Calculate \(\vec{E}\) at the point (1, 0, -2)
Substituting \(x = 1\), \(y = 0\), and \(z = -2\):
\[ \vec{E} = (-4(1), -4(0), -4(-2)) = (-4, 0, 8) \]

### Step 5: Calculate \(\vec{E}\) when \(r = 2\)
When \(r = 2\), we can express \(E\) in terms of \(r\):
\[ V = 2r^2 \implies \vec{E} = -\frac{dV}{dr} \hat{r} \]
Calculating \(\frac{dV}{dr}\):
\[ \frac{dV}{dr} = \frac{d}{dr}(2r^2) = 4r \]
Thus,
\[ \vec{E} = -4r \hat{r} \]
Substituting \(r = 2\):
\[ \vec{E} = -4(2) \hat{r} = -8 \hat{r} \]

### Final Answers
1. \(\vec{E} \text{ at } (1, 0, -2) = (-4, 0, 8)\)
2. \(\vec{E} \text{ at } r = 2 = -8 \hat{r}\)