Two identical circular planar surface charges float in free space (permittivity €0). Thesurface charge density is constant over the disk pso (with units As/m²) for both disks.The disks are separated by distance 'a'. Let's fix the coordinate system such that theupper disk is on the xy plane (z 0) and its center lies on the origin. The lowerplate is on the xy plane (z = −a). The goal is to find the expression for electric scalarpotential at the symmetry axis (z) between the two disks.aPs0 OPs0=Ab(a) Determine the scalar potential V(z) at the z axis in between the two disks. Re-member that you need to integrate the effect of all charge in the disk. (Sincethere are two disks you will need superposition. The cylindrical coordinate sys-tem should be helpful.)Hint:XJ(B² + x2)1/2dxwhere C is constant of integration.= √B² + x² + C

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Two identical circular planar surface charges float in free space (permittivity €0). The
surface charge density is constant over the disk pso (with units As/m²) for both disks.
The disks are separated by distance 'a'. Let's fix the coordinate system such that the
upper disk is on the xy plane (z 0) and its center lies on the origin. The lower
plate is on the xy plane (z = −a). The goal is to find the expression for electric scalar
potential at the symmetry axis (z) between the two disks.
a
Ps0 O
Ps0
=
A
b
(a) Determine the scalar potential V(z) at the z axis in between the two disks. Re-
member that you need to integrate the effect of all charge in the disk. (Since
there are two disks you will need superposition. The cylindrical coordinate sys-
tem should be helpful.)
Hint:
X
J
(B² + x2)1/2dx
where C is constant of integration.
= √B² + x² + C

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Step 1: The potential due to a small element of charge dQ on the disk is given by:
dV = k *dQ/r
where k = 1 / (4+0) is the Coulomb constant, r is the distance from the element of charge to the point
Step 2: Then we have:
dQ = ps0 r dr de
Now we can integrate over the entire disk to get the total potential:
V(z) = JJ dV = k ps0 JJ DA / r
where the integration is over the disk, and the denominator r
Step 3: We can simplify the integral using the hint given in the problem. We have:
V(z) = k ps0 JS dA / √(a² + z²)
= k pso ſо²π de fo^R r dr / √(a² + z²)
where R is the radius of the disk.
Step 4: Using this result and evaluating the outer integral, we get:
V(z) = k ps0 π [√(R² + a² + z²) - √(a² + z²)].
V(z) = k psO π [√(R² + a² + z²) - √(a² + z²)] - k ps◊ í [√(R² + a² + (z + a)²) - √(a² + (z + a)²)]
Simplifying this expression, we get:
V(z) = k psO π [√(R² + a² + z²) - √(a² + z²) - √(R² + a² + (z + a)²) + √(a² +
where k = 1/(40) is the Coulomb constant
+ a)²)]

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