Let's solve the problem step-by-step to find the total charge contained in a non-conducting cylindrical shell with the given charge density.
1. Understanding the Geometry:
- Inner radius of the cylindrical shell: [tex]\( r_{\text{inner}} = \frac{R}{2} \)[/tex]
- Outer radius of the cylindrical shell: [tex]\( r_{\text{outer}} = R \)[/tex]
- Length of the shell: [tex]\( L \)[/tex]
2. Given Charge Density:
- The charge density as a function of radial distance [tex]\( r \)[/tex] is [tex]\( \rho(r) = \frac{3}{r^2} \)[/tex].
3. Formula for Total Charge:
- To find the total charge [tex]\( Q \)[/tex] inside the shell, we need to integrate the charge density over the shell's volume. Since we have a cylindrical shell, we use cylindrical coordinates.
- The volume element in cylindrical coordinates [tex]\( dV \)[/tex] is [tex]\( 2 \pi r L \, dr \)[/tex], where [tex]\( 2 \pi r \)[/tex] is the circumference at a radius [tex]\( r \)[/tex] and [tex]\( L \)[/tex] is the length.
- Therefore, the total charge [tex]\( Q \)[/tex] can be expressed as:
[tex]\[
Q = \int_{r_{\text{inner}}}^{r_{\text{outer}}} \rho(r) \cdot 2 \pi r L \, dr
\][/tex]
4. Substitute the Given Values:
- Substitute [tex]\( \rho(r) = \frac{3}{r^2} \)[/tex], [tex]\( r_{\text{inner}} = \frac{R}{2} \)[/tex], and [tex]\( r_{\text{outer}} = R \)[/tex] into the integral.
[tex]\[
Q = \int_{\frac{R}{2}}^{R} \frac{3}{r^2} \cdot 2 \pi r L \, dr
\][/tex]
- Simplify the integrand:
[tex]\[
Q = 6 \pi L \int_{\frac{R}{2}}^{R} \frac{1}{r} \, dr
\][/tex]
5. Evaluate the Integral:
- The integral [tex]\( \int \frac{1}{r} \, dr \)[/tex] is a standard integral that evaluates to [tex]\( \ln(r) \)[/tex].
[tex]\[
Q = 6 \pi L \left[ \ln(r) \right]_{\frac{R}{2}}^{R}
\][/tex]
- Evaluate the definite integral:
[tex]\[
Q = 6 \pi L \left( \ln(R) - \ln\left(\frac{R}{2}\right) \right)
\][/tex]
- Simplify the logarithmic expression:
[tex]\[
\ln(R) - \ln\left(\frac{R}{2}\right) = \ln\left(\frac{R}{R/2}\right) = \ln(2)
\][/tex]
- Thus, the total charge:
[tex]\[
Q = 6 \pi L \ln(2)
\][/tex]
6. Compare with the Given Options:
- (A) [tex]\( 4 \pi L \ln 2 \)[/tex]
- (B) [tex]\( 6 \pi L \ln 2 \)[/tex]
- (C) [tex]\( 6 \pi R^2 L \)[/tex]
- (D) [tex]\( \frac{2 \pi L}{R^2} \)[/tex]
- (E) Answer is not any of those listed.
From the above calculations, the best expression for the total charge contained in the shell is [tex]\( 6 \pi L \ln(2) \)[/tex]. Thus, the correct answer is (B).
