A uniform charged cube of side L has potential V and electric field E at one corner .if cube of side L/2 is removed find E and V at that point

To tackle the problem of finding the electric field \( E \) and potential \( V \) at the corner of a uniform charged cube after removing a smaller cube from it, we need to break it down step by step. Let's start by understanding the initial conditions and then analyze the changes that occur when we remove the smaller cube.

Initial Setup

Consider a uniform charged cube with side length \( L \) that has a total charge \( Q \). The electric potential \( V \) at a corner of the cube can be calculated using the formula for the potential due to a point charge, integrated over the volume of the cube. The electric field \( E \) at that corner can be derived from the potential or calculated directly from the charge distribution.

Electric Potential and Field of the Full Cube

The potential \( V \) at a corner of the cube is influenced by the entire charge distribution. For a uniformly charged cube, the potential at a corner can be expressed as:

• Potential \( V = k \cdot \frac{Q}{r} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point of interest.

For the electric field \( E \), it can be derived from the potential using the relation:

• Electric Field \( E = -\nabla V \

Removing the Smaller Cube

Now, when we remove a smaller cube of side \( L/2 \) from one corner of the original cube, we need to consider how this affects both the potential and the electric field at that corner. The smaller cube, being uniformly charged, will also have its own charge \( Q' \) proportional to its volume.

Charge of the Smaller Cube

The charge of the smaller cube can be calculated as:

• Charge \( Q' = \rho \cdot \left(\frac{L}{2}\right)^3 = \rho \cdot \frac{L^3}{8} \

where \( \rho \) is the charge density of the original cube.

Effect on Electric Field and Potential

When the smaller cube is removed, the potential \( V \) at the corner will change because we are effectively removing the contribution of the charge \( Q' \) from that point. The new potential \( V' \) can be expressed as:

• New Potential \( V' = V - k \cdot \frac{Q'}{r'} \

Here, \( r' \) is the distance from the remaining charge to the corner, which remains the same since the corner is still at the same location.

Electric Field Calculation

For the electric field \( E' \) at the corner after removing the smaller cube, we can use the principle of superposition. The electric field due to the remaining charge will still be present, but we must also account for the field due to the removed charge:

• New Electric Field \( E' = E - E'_{removed} \

Where \( E'_{removed} \) is the electric field contribution from the removed smaller cube, which can be calculated similarly to how we calculated the original electric field.

Final Expressions

Thus, the new electric field \( E' \) and potential \( V' \) at the corner after removing the smaller cube can be expressed as:

• New Potential \( V' = V - k \cdot \frac{Q'}{r'} \
• New Electric Field \( E' = E - E'_{removed} \

In summary, by carefully analyzing the contributions from both the original and the removed charge distributions, we can derive the new electric field and potential at the corner of the cube. This approach highlights the importance of understanding how charge distributions interact in electrostatics.