Consider a cube of uniform charge density . The ratio of electrostatic potential at the centre of the cube to that at one of the corners of the cube is

Step-by-step solution

Let at the corner of cube potential = V ₀ array l Potential Q Side of cube Q = a ³ array So potential a ³ a Potential a ² Big cube consist of 8 cube At centre of big cube of side 2 a , potential is 8 ~V ₀ Potential at corner of big cube = V ₀ (2)²=4 ~V ₀ Required ratio = 8 ~V ₀ 4 ~V ₀ =2: 1