Eight identical point charges of q coulomb each are placed at the corners of a cube side 0.1m. Calculate electric field at the centre of the cube. Calculate the field at the center when one of the corner charges is removed.

Explanation

To solve the problem of finding the electric field at the center of a cube with point charges at its corners, we will follow these steps:

### Step 1: Understand the Configuration
We have a cube with side length \( a = 0.1 \, \text{m} \) and 8 identical point charges \( q \) placed at each corner of the cube.

### Step 2: Calculate the Electric Field at the Center with All Charges
The electric field \( \vec{E} \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ \vec{E} = k \frac{q}{r^2} \hat{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant.

The distance from each corner of the cube to the center of the cube can be calculated using the formula for the body diagonal of a cube:
\[ r = \frac{a \sqrt{3}}{2} \]
For our cube, this becomes:
\[ r = \frac{0.1 \sqrt{3}}{2} \approx 0.0866 \, \text{m} \]

### Step 3: Determine the Electric Field Contribution from Each Charge
Each charge contributes to the electric field at the center. However, due to symmetry, the contributions from opposite charges will cancel each other out.

Since there are 8 charges symmetrically placed, the net electric field at the center of the cube is:
\[ \vec{E}_{\text{net}} = 0 \]

### Step 4: Calculate the Electric Field at the Center When One Charge is Removed
Now, if we remove one charge from one corner, we are left with 7 charges. The remaining charges will still be symmetrically placed around the center, but now there will be an imbalance.

### Step 5: Determine the Direction of the Electric Field
The electric field at the center due to the remaining 7 charges will point towards the empty corner (where the charge was removed).

### Step 6: Calculate the Electric Field Due to One Charge
The electric field at the center due to one charge \( q \) at the corner is given by:
\[ \vec{E} = k \frac{q}{r^2} \]
Substituting \( r = 0.0866 \, \text{m} \):
\[ \vec{E} = k \frac{q}{(0.0866)^2} \]

### Step 7: Calculate the Total Electric Field Due to 7 Charges
The total electric field at the center due to the 7 charges can be calculated as:
\[ \vec{E}_{\text{total}} = 7 \cdot \vec{E} = 7 \cdot k \frac{q}{(0.0866)^2} \]
This electric field will point towards the empty corner.

### Final Result
The electric field at the center of the cube with all charges is \( 0 \, \text{N/C} \). When one charge is removed, the electric field at the center is directed towards the corner where the charge was removed and has a magnitude of:
\[ \vec{E}_{\text{total}} = 7 \cdot k \frac{q}{(0.0866)^2} \]