A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Answers
Answer:
Side length of a cube is b
∴ diagonal of that cube will be b√3
[ ∵ diagonal = √3 × side length ]
We have to find distance of any vertices from the centre of cube.
e.g., distance of any vertices form the centre of cube = half of diagonal of cube
= √3b/2
Now, potential at centre due to charges placed at vertices of cube
V = 8 × kq/r
[ Because there are 8 vertices in cube ,so total potential at centre of cube is 8 times of potential at centre due to each charge q ]
∴ V = 8kq/√3b/2 = 16kq/√3b
Electric field at the centre of cube becomes zero because a charge at any corner of cube is just equal and opposite to that another charged at diagonally opposite corner of cube. Hence, E = 0
mark brainliest
Answer:
Explanation: electric field is a vector so we have to take direction olso . at centre the direction electric feild due to corner charge will be along the diagonal
Now net EF due to all charges will be zero as the feild produce by one chage will be cancelled by other charge located at its opposite corner.
Potiential is scalar so we just find the potiential due to all the charges and simply add them
Poteintial = KQ/R wher R is the position of the point from charge . Soo here R = (underrooot 3 ) B / 2
Due to all 8 charge potiential = 8 kq/r