Questions: In the figure above a nonuniformly charged rod of length L=4.9 m lies along the x-axis with one end at the origin. The linear charge density (charge per length) is given by λ=(3.8 x 10^-9) x^1.3 (where λ has units of C / m when x has units of meters). Point P is located at the origin (the less densely charged end of the rod). What is the total charge on the rod? (This is given by Q=integral of dq=integral from 0 to L of λ dx.) What is the magnitude of the electric field at point P? (This will be given by E=integral from 0 to L of k λ dx/x^2 where λ is now a function of x instead of a constant.)

In the figure above a nonuniformly charged rod of length L=4.9 m lies along the x-axis with one end at the origin. The linear charge density (charge per length) is given by λ=(3.8 x 10^-9) x^1.3 (where λ has units of C / m when x has units of meters). Point P is located at the origin (the less densely charged end of the rod).
What is the total charge on the rod? (This is given by Q=integral of dq=integral from 0 to L of λ dx.)
What is the magnitude of the electric field at point P? (This will be given by E=integral from 0 to L of k λ dx/x^2 where λ is now a function of x instead of a constant.)

Transcript text: In the figure above a nonuniformly charged rod of length $L=4.9 \mathrm{~m}$ lies along the $x$-axis with one end at the origin. The linear charge density (charge per length) is given by $\lambda=\left(3.8 \times 10^{-9}\right) x^{1.3}$ (where $\lambda$ has units of $\mathrm{C} / \mathrm{m}$ when $x$ has units of meters). Point $P$ is located at the origin (the less densely charged end of the rod). What is the total charge on the rod? (This is given by $Q=\int d q=\int_{0}^{L} \lambda d x$.) What is the magnitude of the electric field at point $P$ ? (This will be given by $E=\int_{0}^{L} k \frac{\lambda d x}{x^{2}}$ where $\lambda$ is now a function of $x$ instead of a constant.)

Solution

What is the total charge on the rod? (This is given by $Q=\int d q=\int_{0}^{L} \lambda d x$.) Calculate the total charge

Given $\lambda = (3.8 \times 10^{-9})x^{1.3}$ and $L = 4.9 \mathrm{~m}$, the total charge $Q$ is given by $$Q = \int_0^L \lambda dx = \int_0^{4.9} (3.8 \times 10^{-9})x^{1.3} dx$$ $$Q = (3.8 \times 10^{-9}) \left[ \frac{x^{2.3}}{2.3} \right]_0^{4.9} = \frac{3.8 \times 10^{-9}}{2.3} (4.9^{2.3})$$ $$Q = \frac{3.8 \times 10^{-9}}{2.3} \times 29.82$$ $$Q = 49.22 \times 10^{-9} \mathrm{C}$$

$\boxed{49.22 \times 10^{-9} \mathrm{C}}$

What is the magnitude of the electric field at point $P$? (This will be given by $E=\int_{0}^{L} k \frac{\lambda d x}{x^{2}}$ where $\lambda$ is now a function of $x$ instead of a constant.) Calculate the electric field

Given $\lambda = (3.8 \times 10^{-9})x^{1.3}$ and $L = 4.9 \mathrm{~m}$, the electric field $E$ at point $P$ is given by $$E = \int_0^L k\frac{\lambda}{x^2} dx = k \int_0^{4.9} \frac{(3.8 \times 10^{-9})x^{1.3}}{x^2} dx$$ $$E = k(3.8 \times 10^{-9}) \int_0^{4.9} x^{-0.7} dx$$ $$E = k(3.8 \times 10^{-9}) \left[ \frac{x^{0.3}}{0.3} \right]_0^{4.9} = k \frac{3.8 \times 10^{-9}}{0.3} (4.9^{0.3})$$ $$E = (8.99 \times 10^9) \frac{3.8 \times 10^{-9}}{0.3}(1.635)$$ $$E = \frac{8.99 \times 3.8 \times 1.635}{0.3} = \frac{55.80}{0.3}$$ $$E = 186 \mathrm{~N/C}$$

$\boxed{186 \mathrm{~N/C}}$

$49.22 \times 10^{-9} \mathrm{C}$ $186 \mathrm{~N/C}$

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