Few q charges are placed at the vertices of the cube. find electrostatic force. Determine the potential and electric field due to this charge array at the center of the cube.
Step 1. Given data:
Le a cube of dimensions shown in the figure below:
Refer to the diagram for the respective dimensions mentioned:
Let the sides of the cube =
Given, that the charge at the vertices =
Step 2. Formula used:
The electric potential (V) at the center of the cube is due to the presence of eight charges at the vertices.
Where = Electric potential. =distance of any of the vertices to the center of the cube.
Step 3. Calculation:
Diagonal of one of the sides of the cube may be given by:
=
Length of the diagonal of the cube
The distance between any one of the vertices and the center of the cube is
The electric potential () at the center of the cube is due to the eight charges at the vertices ( by putting the value of )
Therefore, the potential at the center of the cube will be
=
Step 4. Find electric field intensity:
Hence, the potential at the center of the cube is and the electric field intensity at the center is zero.