To find the electric potential given the electric field and distance, we can use the relationship between electric field (E) and electric potential (V), knowing that electric field is the negative gradient of the electric potential. In a uniform electric field, this simplifies to a linear relationship, where the change in electric potential (∆V) equals the electric field (E) times the distance (∆d). Assuming the potential at the starting point is zero, the formula is:

[ Delta V = E times Delta d ]

Given:

– Electric field (E) = 3 units (assuming SI units, this would be 3 N/C)

– Distance (∆d) = 2 m

Substitute the given values into the formula:

[ Delta V = 3 , text{N/C} times 2 , text{m} = 6 , text{V} ]

So, the electric potential difference is 6 volts.

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The electric flux density (D) and electric field intensity (E) are related through the permittivity of the medium (ε). The relationship is given by the equation:

[ D = epsilon E ]

where:

– ( D ) is the electric flux density (C/m²),

– ( E ) is the electric field intensity (V/m),

– ( epsilon ) is the permittivity of the medium (F/m).

In free space, this is expressed as:

[ D = epsilon_0 E ]

where ( epsilon_0 ) is the permittivity of free space. In a material medium, the relationship accounts for the material’s relative permittivity (εr) as well:

[ D = epsilon_0 epsilon_r E ]

This means that the electric flux density is proportional to the electric field intensity, with the permittivity acting as the proportionality constant.

To find the electric flux density ( D ) at ( R = 6m ) due to the three charged cylindrical sheets, we can use Gauss’s Law for electric fields. The key steps are:

1. Identify the charge density: The charge densities given are:

– For the first sheet at ( R = 2m ), ( sigma_1 = 5 , text{C/m}^2 )

– For the second sheet at ( R = 4m ), ( sigma_2 = -2 , text{C/m}^2 )

– For the third sheet at ( R = 5m ), ( sigma_3 = -3 , text{C/m}^2 )

2. Calculate the contribution to the flux density ( D ) at ( R = 6m ) from each sheet. The formula for the electric flux density ( D ) due to an infinite plane sheet of charge is:

[

D = frac{sigma}{2} quad text{(points away from positively charged sheet)}

]

– For the first sheet:

[

D_1 = frac{5}{2} = 2.5 , text{C/m}^2 quad (text{at } R > 2m)

]

– For the second sheet:

To find the electric flux density (D) at R = 4.5 m due to the three charged cylindrical sheets, we can use Gauss’s law for electricity, which relates the electric flux density to the surface charge density.

The electric flux density D due to a uniformly charged infinite cylindrical sheet is given by the equation:

[

D = frac{sigma}{2} hat{n}

]

where σ is the surface charge density and (hat{n}) is the unit normal vector pointing away from the sheet.

Given:
1. At R = 2 m, σ = 5 C/m²
2. At R = 4 m, σ = -2 C/m²
3. At R = 5 m, σ = -3 C/m²

Now, we calculate the contributions to D at R = 4.5 m from all three sheets.

1. Contribution from the sheet at R = 2 m:

[

D_1 = frac{5}{2} = 2.5 , text{(outward)}

]

2. Contribution from the sheet at R = 4 m:

[

D_2 = frac{-2}{2} = -1 , text{(inward)}

]

3. Contribution from the sheet at R = 5 m:

[

D_