Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.
A small positive point charge of 8 mC moves from a distance of 9 cm to a distance of 17 cm.
How much work is done by the electric field?
Hint: The electric field for a long charged line is:
Express the result in the unit mJ and to three significant figures.
The Equation is E line = 1/(4piE0)(2lamda/r)
Since you have not provided the "hint", I used Gauss' Law to come up with
2*pi*R*L*E = L*q/eo
where eo is the coulomb's law constant "epsilon-zero" = 8.89^10^-12 N*m^2/C^2
and q is the charge per unit length
E(R) = q/(2 pi eo R)
Multiply that by Q = 8 mC and integrate with dR from R = 0.09 to 0.17 m
There should be a ln(17/8) term in the answer.
The work done will be positive since Q is moving in the opposite direction from the attractive force
The hint is
E line = 1/(4piE0)(2lamda/r)
Are the answers the same even if I use Gauss' Law to come up with 2*pi*R*L*E = L*q/eo
Isn't the work negative, because this is a positive charge moving in the opposite direction of the electric field?
Positive work must be done in order to make a positive charge go opposite the direction of the electric field.
F of Electric field----> <-- + F of Positive charge
(If you did negative work, the positive charge would move in the direction of the field.)
To find the work done by the electric field on the small positive point charge, we can use the formula:
Work = -ΔPE = -qΔV,
where ΔPE is the change in potential energy, q is the charge, and ΔV is the change in voltage.
First, let's determine the electric field created by the long charged line. The hint provides the electric field formula, which is:
E = λ / (2πε₀r),
where λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the charged line.
Given that the linear charge density is -2.50 nC/m, we can convert it to coulombs per meter (C/m):
λ = -2.50 nC/m = -2.50 × 10^(-9) C/m.
The permittivity of free space is ε₀ = 8.85 × 10^(-12) C²/(N·m²).
Now, to find the electric field at each distance, we substitute r = 9 cm and r = 17 cm into the formula:
For r = 9 cm = 0.09 m:
E₁ = (-2.50 × 10^(-9) C/m) / (2π(8.85 × 10^(-12) C²/(N·m²))(0.09 m),
For r = 17 cm = 0.17 m:
E₂ = (-2.50 × 10^(-9) C/m) / (2π(8.85 × 10^(-12) C²/(N·m²))(0.17 m).
Calculate both values of E₁ and E₂.
Now, the change in electric potential energy (ΔPE) is given by the equation:
ΔPE = qΔV,
where ΔV is the change in voltage. In this case, we can calculate ΔV by using the formula:
ΔV = -E₁Δr,
where Δr is the change in distance, which is given as 0.17 m - 0.09 m.
Now, calculate ΔV by substituting the values.
Finally, substitute the values of q = 8 mC and ΔV into the equation for work:
Work = -qΔV.
Calculate the value of work and express it in millijoules (mJ) rounded to three significant figures.