Electrostatic potential of a cube with charge density

The potential at the center of the larger cube is the sum of the potentials at the center of each smaller cube. The potential at the corner of the larger cube is the sum of the potentials at the corners of the smaller cubes. Therefore, the ratio of the electrostatic potential at the centre of the cube to that of one of the corners is 1:8.
  • #1
Yashasvi Grover
9
0

Homework Statement



Consider a uniformly charged cube with uniform charge density ρ.The ratio of electrostatic potential at the centre of the cube to that of one of the corners of the cube is?
A hint on how to approach the problem's solution would be appreciated.(whether to use gauss law or not etc.)

Homework Equations


Phi=E.A and
ΔV=Workdone/charge=-E.D

The Attempt at a Solution


I tried to use Gauss law and symmetry argument to find electric field inside the cube,but failed.Help would be appreciated.
 
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  • #2
Yashasvi Grover said:

Homework Statement



Consider a uniformly charged cube with uniform charge density ρ.The ratio of electrostatic potential at the centre of the cube to that of one of the corners of the cube is?
A hint on how to approach the problem's solution would be appreciated.(whether to use gauss law or not etc.)
2. Homework Equations
Phi=E.A and
ΔV=Workdone/charge=-E.D

The Attempt at a Solution


I tried to use Gauss law and symmetry argument to find electric field inside the cube,but failed.Help would be appreciated.
Consider it made of eight half-sized cubes.
 

FAQ: Electrostatic potential of a cube with charge density

1. What is meant by electrostatic potential?

The electrostatic potential is a measure of the potential energy per unit charge at a given point in an electric field. It is a scalar quantity and is measured in volts (V).

2. How is the electrostatic potential of a cube with charge density calculated?

The electrostatic potential of a cube with charge density can be calculated by using the formula V = k * q / r, where V is the potential, k is the Coulomb's constant, q is the charge, and r is the distance from the charge to the point where the potential is being measured.

3. What is the relationship between electrostatic potential and electric field?

The electric field is the negative gradient of the electrostatic potential. In other words, the electric field is the rate of change of the electrostatic potential with respect to distance. This relationship is given by E = -∇V, where E is the electric field, ∇ is the gradient operator, and V is the electrostatic potential.

4. How does the electrostatic potential change as the distance from the charge increases?

As the distance from the charge increases, the electrostatic potential decreases. This is because the electric field strength decreases with distance, and the electrostatic potential is directly proportional to the electric field strength. Therefore, the further away from the charge, the lower the electrostatic potential will be.

5. Can the electrostatic potential of a cube with charge density be negative?

Yes, the electrostatic potential of a cube with charge density can be negative. This indicates that the potential energy at that point is negative, meaning that the charge would experience a force in the opposite direction of the electric field. This is commonly seen in regions where there are multiple charges with different signs, resulting in a net negative potential energy.

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