A long cylindrical insulator has a uniform charge density of 1.0 #C/m3 and a radius of
9.0 cm.
a) What is the electric field inside the insulator at a distance of 4 cm?
b) What is the electric field at 17 cm?
c) How much work must you do to bring a q = 0.05 #C test charge from 17 cm
to 4 cm?
# means micro
Use Gauss' Law inside and outside the cylinder
a) Inside a uniformly charged cylinder of radius r < 0.09 m, there will be
Q = 10^-6 * pi*r^2*L Coulombs
Gauss' Law says, for this geometry, that
2*pi*r*L*E(r) = Q/epsilon
= 10^-6 * pi*r^2*L
Therefore
E(r) = 10^-6*r/(2*epsilon)
= 2260 Volt/m at r = 0.04 m
For epsilon, I used 8.85*10^-12 C^2/(N*m^2)
(b) Let R = 0.09 m be the solid cylinder radius. r = 17 cm is outside the solid cylinder. The charge inside an outer cylinder of radius r > R is
Q = 10^-6*pi*R^2*L
and the E- field at r > R is again given by Gauss' law
2 pi*r*L*E(r) = 10^-6*pi*R^2*L
E(r) = 10^-6 R^2/(2*r*epsilon)
= 5085*(R/r) Volt/m
c) integrate
(test charge)*E*dr from r = 17 to r = 4 cm. You will have to do in in two parts, inside and outside r = R, since the formulas are different in the two regions.
To solve these questions, we can use the formula for the electric field inside a uniformly charged cylinder:
E = (ρ * r) / (2ε₀)
where:
- E is the electric field strength
- ρ is the charge density of the cylinder
- r is the distance from the center of the cylinder
- ε₀ is the electric constant (ε₀ ≈ 8.854 x 10⁻¹² C²/Nm²)
Given:
- ρ = 1.0 C/m³
- r₁ = 4 cm = 0.04 m
- r₂ = 17 cm = 0.17 m
- q = 0.05 C (test charge)
Now we can calculate each part of the problem step by step:
a) What is the electric field inside the insulator at a distance of 4 cm?
Using the formula,
E₁ = (ρ * r₁) / (2ε₀)
E₁ = (1.0 * 0.04) / (2 * 8.854 x 10⁻¹²)
E₁ ≈ 2.27 x 10⁸ N/C
b) What is the electric field at 17 cm?
Using the formula,
E₂ = (ρ * r₂) / (2ε₀)
E₂ = (1.0 * 0.17) / (2 * 8.854 x 10⁻¹²)
E₂ ≈ 4.84 x 10⁷ N/C
c) How much work must you do to bring the test charge from 17 cm to 4 cm?
The electric potential difference between two points is given by:
ΔV = - ∫E * ds
The work done is given by:
W = q * ΔV
To find the work, we need to integrate the electric field between the two points.
W = q * (V₁ - V₂)
First, let's find the potential at each point:
V₁ = - ∫E * ds from r₁ to r₂
V₁ = - ∫[ (ρ * r) / (2ε₀) ] * dr from r₁ to r₂
V₁ = - ∫(1.0 * r) / (2 * 8.854 x 10⁻¹²) * dr from 0.04 to 0.17
Evaluating this integral, we get V₁ ≈ -6.70 x 10⁶ V
Similarly, V₂ = - ∫[ (ρ * r) / (2ε₀) ] * dr from 0.17 to 0.04
Evaluating this integral, we get V₂ ≈ -2.14 x 10⁶ V
Now we can calculate the work:
W = q * (V₁ - V₂)
W = 0.05 * [ (-6.70 x 10⁶) - (-2.14 x 10⁶) ]
W = 0.05 * (-4.56 x 10⁶)
W ≈ -2.28 x 10⁵ J (negative because the work is done against the electric field)
To find the electric field inside a long cylindrical insulator with a uniform charge density, we can use the formula for electric field intensity due to a uniformly charged cylinder:
E = ρ * r / (2 * ε₀)
where:
- E is the electric field intensity
- ρ is the charge density
- r is the distance from the center of the cylinder
- ε₀ is the permittivity of free space
a) To find the electric field inside the insulator at a distance of 4 cm (0.04 m), we can substitute the values into the formula:
E = (1.0 #C/m³) * (0.04 m) / (2 * ε₀)
The value of ε₀ is 8.85 x 10⁻¹² C² / Nm². By substituting this value, we can find the electric field.
E = (1.0 #C/m³) * (0.04 m) / (2 * 8.85 x 10⁻¹² C² / Nm²)
b) To find the electric field at 17 cm (0.17 m), we follow the same process:
E = (1.0 #C/m³) * (0.17 m) / (2 * ε₀)
c) To find the work required to bring a test charge from 17 cm to 4 cm, we need to calculate the electric potential difference between these two points first. The formula for electric potential due to a uniformly charged cylinder is:
V = ρ * r² / (4 * ε₀)
We can calculate the potential difference as:
ΔV = V₁ - V₂ = (ρ * r₁² / (4 * ε₀)) - (ρ * r₂² / (4 * ε₀))
Finally, we can calculate the work done as:
W = q * ΔV
where q is the test charge in coulombs.
By following these steps and substituting the given values, we can find the answers to the given questions.