To calculate the electric potential at a distance of 6.0 m from an infinite line charge, we first need to understand how the electric potential due to a line charge works. The formula for the electric potential \( V \) at a distance \( r \) from an infinite line charge with linear charge density \( \lambda \) is given by:
Electric Potential Formula
The formula is expressed as:
V(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln\left(\frac{r}{r_0}\right)
In this equation:
- V(r) is the electric potential at distance r.
- λ is the linear charge density.
- ε0 is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).
- r is the distance from the line charge where we want to find the potential.
- r0 is the reference distance where the potential is defined to be zero.
Given Values
From your question, we know:
- Linear charge density, \( \lambda = 0.3 \times 10^{-6} \, \text{C/m} \)
- Reference distance, \( r_0 = 5.0 \, \text{m} \)
- Distance where we want to find the potential, \( r = 6.0 \, \text{m} \)
Calculating the Potential
Now, we can plug in the values into the formula. First, we need to calculate the potential at 6.0 m:
V(6.0) = -\frac{0.3 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12}} \ln\left(\frac{6.0}{5.0}\right)
Calculating the constant:
-\frac{0.3 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12}} \approx -5.37 \times 10^{4} \, \text{V}
Next, we calculate the logarithmic term:
\ln\left(\frac{6.0}{5.0}\right) \approx \ln(1.2) \approx 0.1823
Now, substituting back into the potential equation:
V(6.0) = -5.37 \times 10^{4} \times 0.1823 \approx -9800 \, \text{V}
Final Result
Thus, the electric potential at a distance of 6.0 m from the infinite line charge, assuming the potential at 5.0 m is zero, is approximately:
V(6.0) \approx -9800 \, \text{V}
This negative value indicates that the potential is lower at 6.0 m compared to the reference point at 5.0 m, which is consistent with the behavior of electric potentials around positive line charges.
