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Give step-by-step solution with explanation and final answer:A charged nonconducting rod, with a length of 1.11 m and a cross-sectional area of 3.25 cm? lies along the positive side of an x axis with one end at the origin. The volume charge density p is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if p is (a) uniform, with a value of -3.30 uC/m® and (b) nonuniform, with a value given by p = bx2, where b=-2.53uC/m5?

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Give step-by-step solution with explanation and final answer:Uploaded ImageA charged nonconducting rod, with a length of 1.11 m and a cross-sectional area of 3.25 cm? lies along the positive side of an x axis with one end at the origin. The volume charge density p is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if p is (a) uniform, with a value of -3.30 uC/m® and (b) nonuniform, with a value given by p = bx2, where b=-2.53uC/m5?

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Asked by: Rahul
Created at: 2026-01-17 00:56:11

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Let's break down the problem step by step. --- ## **Given Data** - **Length of rod,** \( L = 1.11 \) m - **Cross-sectional area,** \( A = 3.25 \) cm² \( = 3.25 \times 10^{-4} \) m² - **Volume charge density,** \( \rho \) - **Elementary charge,** \( e = 1.602 \times 10^{-19} \) C --- ## **(a) Uniform Charge Density** ### **Step 1: Convert \(\rho\) to SI units** \[ \rho = -3.30~\mu\text{C/m}^3 = -3.30 \times 10^{-6}~\text{C/m}^3 \] ### **Step 2: Calculate the total charge \( Q \)** \[ Q = \rho \times \text{Volume} = \rho \times (A \times L) \] \[ Q = (-3.30 \times 10^{-6}) \times (3.25 \times 10^{-4}) \times (1.11) \] Calculate inside the parentheses: \[ 3.25 \times 10^{-4} \times 1.11 = 3.6075 \times 10^{-4} \] So, \[ Q = -3.30 \times 10^{-6} \times 3.6075 \times 10^{-4} \] \[ Q = -1.190475 \times 10^{-9}~\text{C} \] ### **Step 3: Find the number of excess electrons, \( n \)** \[ n = \frac{|Q|}{e} \] \[ n = \frac{1.190475 \times 10^{-9}}{1.602 \times 10^{-19}} \] \[ n \approx 7.43 \times 10^{9} \] --- ## **(b) Nonuniform Charge Density, \(\rho = bx^2\), \( b = -2.53~\mu\text{C/m}^5 \)** ### **Step 1: Convert \( b \) to SI units** \[ b = -2.53~\mu\text{C/m}^5 = -2.53 \times 10^{-6}~\text{C/m}^5 \] ### **Step 2: Write the expression for total charge** \[ dQ = \rho \cdot dV = bx^2 \cdot (A dx) \] \[ Q = \int_^L b x^2 A dx = bA \int_^L x^2 dx = bA \left[ \frac{x^3}{3} \right]_^L = bA \frac{L^3}{3} \] ### **Step 3: Substitute values** \[ Q = (-2.53 \times 10^{-6}) \times (3.25 \times 10^{-4}) \times \frac{(1.11)^3}{3} \] Calculate \( (1.11)^3 = 1.367631 \) \[ Q = (-2.53 \times 10^{-6}) \times (3.25 \times 10^{-4}) \times \frac{1.367631}{3} \] \[ = (-2.53 \times 10^{-6}) \times (3.25 \times 10^{-4}) \times .455877 \] \[ = (-2.53 \times 10^{-6}) \times 1.4836 \times 10^{-4} \] \[ = -3.7535 \times 10^{-10}~\text{C} \] ### **Step 4: Find the number of excess electrons** \[ n = \frac{|Q|}{e} \] \[ n = \frac{3.7535 \times 10^{-10}}{1.602 \times 10^{-19}} \] \[ n \approx 2.34 \times 10^{9} \] --- ## **Final Answers** ### **(a) Uniform charge density:** \[ \boxed{7.43 \times 10^9~\text{excess electrons}} \] ### **(b) Nonuniform charge density (\(\rho = bx^2\)):** \[ \boxed{2.34 \times 10^9~\text{excess electrons}} \]

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