A charged line extends from y = 2.50 cm to y = -2.50 cm. The total charge distributed uniformly along the line is -9.00 nC. At what distance x does the result for the infinite line of charge differ by 1.0% from that for the finite line?
To determine the distance at which the result for the infinite line of charge differs by 1.0% from that for the finite line, we can use the principle of superposition.
The electric field produced by a finite line of charge is given by:
E_finite = (k * λ) / (2πε₀) * [(a+y₁) / √(a² + y₁²)] + [(a-y₂) / √(a² + y₂²)]
where k is Coulomb's constant (9.0 x 10^9 Nm²/C²), λ is the charge density (total charge divided by the length of the line), ε₀ is the permittivity of free space (8.85 x 10^-12 C²/Nm²), a is the distance from the charged line, and y₁ and y₂ represent the y-coordinates of the endpoints of the line.
The electric field produced by an infinite line of charge is given by:
E_infinite = (k * λ) / (2πε₀a)
To find the distance where the results differ by 1.0%, we need to calculate the electric fields for both cases and then solve for the value of a that satisfies the condition (E_finite - E_infinite) / E_infinite = 0.01.
Let's calculate the electric fields for both cases:
For the finite line of charge:
λ = total charge / length of the line
= -9.00 nC / (2 * 2.50 cm)
= -3.60 nC/cm
For the infinite line of charge:
a = distance at which we want to compare the results
Now let's plug in these values into the formulas and solve for the value of a:
(E_finite - E_infinite) / E_infinite = 0.01
[(k * λ) / (2πε₀) * [(a+y₁) / √(a² + y₁²)] + [(a-y₂) / √(a² + y₂²)] - (k * λ) / (2πε₀a)] / ((k * λ) / (2πε₀a)) = 0.01
Solving this equation will give us the value of a, which represents the distance at which the results differ by 1.0% in the electric field.