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 12 mC moves from a distance of 7 cm to a distance of 18 cm.
In mJ, how much work is done by the electric field?
Hint: The electric field for a long charged line is
The electric field for a long charged line is given by the equation:
E = (λ / 2πε₀r)
Where:
E is the electric field
λ is the linear charge density
ε₀ is the permittivity of free space
r is the distance from the charged line
First, let's calculate the electric field at the initial distance of 7 cm:
E_initial = (-2.50 nC/m) / (2π * ε₀ * 0.07 m)
Next, let's calculate the electric field at the final distance of 18 cm:
E_final = (-2.50 nC/m) / (2π * ε₀ * 0.18 m)
Now, let's calculate the work done by the electric field when the positive point charge moves from the initial distance to the final distance using the formula:
Work = q * (ΔV)
Where:
Work is the work done
q is the charge
ΔV is the change in electric potential
Since the charge is positive (+12 mC), the sign of the work done will be positive:
Work = (12 mC) * (ΔV)
To find the change in electric potential, we can rearrange the formula as:
ΔV = -∫(E * ds)
Where:
ΔV is the change in electric potential
E is the electric field
ds is the differential distance along the path
By integrating the equation, we can find the change in electric potential and then calculate the work done.
Note: The integration of the equation will depend on the shape of the path the charge takes. If you provide more information about the path, I can assist you further.
To calculate the work done by the electric field, we need to find the electric force on the small positive charge and then calculate the work done as the charge moves between the two distances.
The electric field for a long charged line is given by:
E = (λ / (2πε₀r)),
where E is the electric field, λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the line.
In this case, the linear charge density is given as -2.50 nC/m, which can be converted to -2.50 × 10⁻⁹ C/m.
Given that the initial distance is 7 cm (or 0.07 m) and the final distance is 18 cm (or 0.18 m), we can calculate the electric fields at these distances.
Initial electric field (E₁) at 7 cm:
E₁ = (-2.50 × 10⁻⁹ C/m) / (2πε₀(0.07 m))
Final electric field (E₂) at 18 cm:
E₂ = (-2.50 × 10⁻⁹ C/m) / (2πε₀(0.18 m))
Now we can calculate the work done by the electric field using the formula:
Work = (q * ΔV),
where q is the charge and ΔV is the change in voltage (potential difference) as the charge moves between the two distances.
Since the work is given in millijoules (mJ), we need to convert the charge from coulombs (C) to millicoulombs (mC).
The charge is given as 12 mC.
Let's calculate the work done:
Work = (12 × 10⁻³ C) * (ΔV)
To find ΔV, we can use the formula:
ΔV = E₂ - E₁
Finally, substitute the values and calculate the work done to find the final answer in mJ.