Question

find the total charge contained in a cone defined by r ≤ 2 m and 0 ≤ ≤ 4⁄ , given that volume charge density = 1022 /3.

Answer 1

To find the total charge contained in a cone defined by a given charge density function, we need to integrate the charge density function over its volume. By expressing the volume and charge density in terms of the variable r, we can compute the total charge by integrating the charge density function over the limits of the cone.

Step-by-step explanation:

To find the total charge contained in a cone, we need to integrate the charge density function over its volume. The charge density function is given as p(r) = (1022/3) / r^2.

1. First, we need to express the volume and charge density in terms of the variable r. Since the cone has a radius limit of r ≤ 2m, we can express its volume as V = (1/3) * π * r^2 * h, where h is the height of the cone.
2. Next, we need to express the charge density in terms of radius r. Since we have p(r) = (1022/3) / r^2, we substitute r with x and multiply by the factor x^2 / x^2, which gives us (1022/3) * (x^2 / x^2) = (1022/3) * (x^2 / 4^2).
3. To find the total charge, we need to integrate the charge density function over the volume of the cone. Therefore, we integrate (1022/3) * (x^2 / 4^2) over the limits of r = 0 to r = 2.
4. Integrating the function with respect to x, we get Q = ∫[(1022/3) * (x^2 / 4^2)]dx, where the limits of integration are from 0 to 2.
5. Solving the integral, we have Q = (1022/3) * [x^3 / (4^2 * 3)] evaluated from 0 to 2.
6. Simplifying the expression, we find Q = (1022/3) * [2^3 / (4^2 * 3)] - (1022/3) * [0^3 / (4^2 * 3)].
7. Finally, simplifying further, we get Q = (1022/3) * [8 / (16 * 3)].

Therefore, the total charge contained in the cone defined by r ≤ 2m and 0 ≤ θ ≤ 4π/3 is (1022/3) * [8 / (16 * 3)].