A cube of side b has a charge q at each of its vertices The potential due to - Physics - Electrostatic Potential And Capacitance

Question

A   c u b e   o f   s i d e   b   h a s
  a   c h a r g e   q   a t   e a c h
  o f   i t s   v e r t i c e s   T h e  
  p o t e n t i a l   d u e   t o   t h i s
  c h a r g e   a r r a y   a t   t h e  
c e n t r e   o f   t h e   c u b e   i s ( a )
  z e r o ( b )   4 q 3 π ε 0 b ( c )   q
π ε 0 b ( d )   4 q π ε 0 b

Decoder_2 · Accepted Answer

Dear student, Length of the side of a cube = b Charge at each of its vertices = q A cube of side b is shown in the following figure

d = Diagonal of one of the six faces of the cube

l = Length of the diagonal of the cube

The electric potential (V) at the center of the cube is due to the presence of eight charges at the vertices

Therefore, the potential at the center of the cube is

Regards

A c u b e o f s i d e b h a s a c h a r g e q a t e a c h o f i t s v e r t i c e s . T h e p o t e n t i a l d u e t o t h i s c h a r g e a r r a y a t t h e c e n t r e o f t h e c u b e i s ( a ) z e r o ( b ) 4 q 3 π ε 0 b ( c ) q π ε 0 b ( d ) 4 q π ε 0 b

$A c u b e o f s i d e b h a s a c h a r g e q a t e a c h o f i t s v e r t i c e s . T h e p o t e n t i a l d u e t o t h i s c h a r g e a r r a y a t t h e c e n t r e o f t h e c u b e i s (a) z e r o (b) \frac{4 q}{\sqrt{3} π ε_{0} b} (c) \frac{q}{π ε_{0} b} (d) \frac{4 q}{π ε_{0} b}$