To find the potential difference due to a point charge, we can use the formula for electric potential ( V ) due to a point charge ( Q ):

[

V = frac{k cdot Q}{r}

]

where ( k ) is Coulomb’s constant (( k approx 8.99 times 10^9 , text{N m}^2/text{C}^2 )), ( Q ) is the charge, and ( r ) is the distance from the charge to the point where the potential is being calculated.

1. Location of the charge: ( (2, 3, 3) ) m
2. Charge ( Q ): ( 0.4 , text{nC} = 0.4 times 10^{-9} , text{C} )
3. Location A (where we calculate the potential): ( (2, 3, 3) ) m (same as the charge location)
4. Location B: ( (-2, 3, 3) ) m

### Step 1: Calculate potential at Point A

For point A, the distance ( r_A ) from the charge is 0, since it is located at the same point as the charge.

[

V_A = frac{k cdot Q}{r_A} quad

To calculate the electric potential ( V ) at a point in space due to a point charge, we can use the formula:

[

V = frac{k cdot Q}{r}

]

Where:

– ( V ) is the electric potential,

– ( k ) is Coulomb’s constant, approximately ( 8.99 times 10^9 , text{N m}^2/text{C}^2 ),

– ( Q ) is the charge (in coulombs),

– ( r ) is the distance from the charge to the point where the potential is being calculated (in meters).

In this case:

– The point charge ( Q = 2 , text{nC} = 2 times 10^{-9} , text{C} ).

– The point where we want to calculate the potential is at ( (1,0,0) ) meters, which is 1 meter away from the origin where the charge is located.

Now, substituting the values into the formula:

1. Calculate ( r ):

[

r = 1 , text{m}

]

2. Substitute ( k ), ( Q ), and ( r ) into the formula:

[

V = frac{(8.99 times 10^9) cdot (2 times 10^{-9})}{

To calculate the total electric flux (Φ) through the surface of a sphere, you can use Gauss’s law, which states that the total electric flux through a closed surface is equal to the electric flux density (D) multiplied by the surface area of the sphere (A).

1. Calculate the Surface Area of the Sphere (A):

The formula for the surface area of a sphere is:

[

A = 4pi r^2

]

Given the radius ( r = frac{1}{4pi} , text{m} ):

[

A = 4pi left(frac{1}{4pi}right)^2 = 4pi cdot frac{1}{16pi^2} = frac{1}{4pi}

]

2. Calculate the Total Electric Flux (Φ):

The total electric flux is given by:

[

Φ = D cdot A

]

where ( D = 16pi ) (electric flux density). Now substituting the values:

[

Φ = 16pi cdot frac{1}{4pi} = 4

]

So, the total flux is 4 units.

To find the electric flux density ( mathbf{D} ) due to a uniform surface charge density ( sigma ) located in the plane ( z = 2 ), we can use Gauss’s law.

The electric flux density ( mathbf{D} ) near an infinite plane sheet of charge is given by:

[

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

]

where ( hat{n} ) is the unit normal vector pointing away from the surface. The surface charge density ( sigma = 2 , mu C/m^2 = 2 times 10^{-6} , C/m^2 ).

The normal vector ( hat{n} ) points in the positive ( z )-direction since the sheet is placed at ( z = 2 ) and we are considering points above the sheet.

Calculating the flux density ( mathbf{D} ):

[

mathbf{D} = frac{2 times 10^{-6} , C/m^2}{2} hat{k} = 10^{-6} , C/m^2 hat{k}

]

Since the point ( P(1, 1, 1) ) is below the plane at ( z = 2 ), we should consider the contribution of the plane charge in the ( z = 1 ) direction (down