A charged nonconducting rod, with a length of 3.83 m and a cross-sectional area of 5.69 cm2, lies along the positive side of an x axis with one end at the origin. The volume charge density ρ is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if ρ is (a) uniform, with a value of -3.76 µC/m3, and (b) nonuniform, with a value given by ρ = bx2, where b = -2.58 µC/m5?
To find the number of excess electrons on the rod, we need to first calculate the total charge on the rod using the given charge density.
(a) For a uniform charge density, the charge density remains constant throughout the rod. The formula to calculate the charge is given by:
Q = ρ * V
where Q is the total charge, ρ is the charge density, and V is the volume of the rod.
The volume of the rod can be calculated using the length and cross-sectional area:
V = L * A
where L is the length and A is the cross-sectional area.
Substituting the given values:
V = 3.83 m * (5.69 cm^2 * 10^-4 m^2/cm^2) [Converting cm^2 to m^2]
V = 0.0218 m^3
Now, we can calculate the total charge:
Q = -3.76 µC/m^3 * 0.0218 m^3 [Converting µC to C]
Q ≈ -0.0824 C
The charge is negative, which indicates the presence of electrons on the rod.
The elementary charge (e) of an electron is approximately -1.602 x 10^-19 C. To find the number of excess electrons, we can divide the total charge by the elementary charge:
Number of excess electrons = Q / e
Number of excess electrons ≈ -0.0824 C / (-1.602 x 10^-19 C)
Number of excess electrons ≈ 5.14 x 10^17 electrons
Therefore, for a uniform charge density, there are approximately 5.14 x 10^17 excess electrons on the rod.
(b) For a nonuniform charge density given by ρ = bx^2, where b = -2.58 µC/m^5, we need to integrate the charge density over the volume of the rod to find the total charge.
The volume charge density can be written as ρ = dq / dV, where dq is an infinitesimal charge element and dV is an infinitesimal volume element.
Since the charge density varies with x, we can express dq in terms of dx and rewrite the equation as ρ = dq / dx * dx / dV.
dq = ρ * dx * dV
Now, we can integrate this equation to find the total charge Q:
Q = ∫ dq = ∫ ρ * dx * dV
Since the cross-sectional area A is constant, we can write dV = A * dx.
Q = ∫ ρ * A * dx
Q = A * ∫ ρ * dx
Substituting the value of ρ = bx^2:
Q = A * ∫ b * x^2 * dx
Q = b * A * ∫ x^2 * dx
Q = b * A * [x^3 / 3]
To find the x limits of integration, we need to determine the range of x values along the rod. Given that one end of the rod is at the origin (x = 0), and the length of the rod is given as 3.83 m, the other end of the rod will be at x = 3.83 m.
Substituting the limits and the given values:
Q = b * A * [(3.83 m)^3 / 3]
Q = -2.58 µC/m^5 * 5.69 cm^2 * 10^-4 m^2/cm^2 * [(3.83 m)^3 / 3] [Converting µC to C and cm^2 to m^2]
Q ≈ -2.72 x 10^-14 C
Again, the negative sign indicates excess electrons on the rod.
Number of excess electrons = Q / e
Number of excess electrons ≈ -2.72 x 10^-14 C / (-1.602 x 10^-19 C)
Number of excess electrons ≈ 1.70 x 10^5 electrons
Therefore, for a nonuniform charge density, there are approximately 1.70 x 10^5 excess electrons on the rod.