Question

two long, charged, thin-walled, concentric cylindrical shells have radii of ( r_i = 3.0 , \text{cm} ) and ( r_o = 6.0 , \text{cm} ). the charge per unit length is ( -7.0 \times 10^{-6} , \text{c/m} ) on the inner shell and ( 5.0 \times 10^{-6} , \text{c/m} ) on the outer shell. what is the magnitude of the electric field at a radius ( r = 4.0 , \text{cm} )? options: ( -6.3 \times 10^6 , \text{n/c} ), ( 7.9 \times 10^7 , \text{n/c} ), ( 2.3 \times 10^6 , \text{n/c} ), ( 3.1 \times 10^6 , \text{n/c} ).

Explanation:

Step1: Recall Gauss's Law for Cylindrical Symmetry

For a long, charged cylindrical shell, the electric field at a distance \( r \) from the axis (where \( r_i < r < r_o \)) is given by \( E = \frac{\lambda_{enclosed}}{2\pi\epsilon_0 r} \), where \( \lambda_{enclosed} \) is the charge per unit length enclosed, \( \epsilon_0 = 8.85\times 10^{-12} \, \text{C}^2/\text{N·m}^2 \), and \( r \) is the radial distance. Here, \( r = 4.0 \, \text{cm} = 0.04 \, \text{m} \), and the enclosed charge per unit length is \( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \) (since we are between the inner and outer shells, only the inner shell's charge is enclosed).

Step2: Substitute Values into the Formula

First, calculate the magnitude (we can take absolute value for magnitude, but let's follow the formula). The formula for the magnitude of the electric field (since we want magnitude, we consider the absolute value of \( \lambda_{enclosed} \)):

\( |E| = \frac{|\lambda_{enclosed}|}{2\pi\epsilon_0 r} \)

Substitute \( |\lambda_{enclosed}| = 7.0\times 10^{-6} \, \text{C/m} \), \( \epsilon_0 = 8.85\times 10^{-12} \, \text{C}^2/\text{N·m}^2 \), and \( r = 0.04 \, \text{m} \):

\( |E| = \frac{7.0\times 10^{-6}}{2\pi\times 8.85\times 10^{-12} \times 0.04} \)

First, calculate the denominator: \( 2\pi\times 8.85\times 10^{-12} \times 0.04 \approx 2\times 3.1416\times 8.85\times 10^{-12} \times 0.04 \approx 2.224\times 10^{-12} \)

Then, \( |E| = \frac{7.0\times 10^{-6}}{2.224\times 10^{-12}} \approx 3.147\times 10^{6} \, \text{N/C} \approx 3.1\times 10^{6} \, \text{N/C} \)? Wait, no, wait, maybe I made a mistake. Wait, let's recalculate the denominator:

\( 2\pi\epsilon_0 r = 2\times\pi\times 8.85\times 10^{-12} \times 0.04 \)

\( 2\times 3.1416 = 6.2832 \)

\( 6.2832\times 8.85\times 10^{-12} = 55.60632\times 10^{-12} = 5.560632\times 10^{-11} \)

\( 5.560632\times 10^{-11} \times 0.04 = 2.2242528\times 10^{-12} \)

Then numerator: \( 7.0\times 10^{-6} \)

So \( E = \frac{7.0\times 10^{-6}}{2.2242528\times 10^{-12}} \approx 3.147\times 10^{6} \, \text{N/C} \approx 3.1\times 10^{6} \, \text{N/C} \)? Wait, but let's check the options. Wait, maybe I messed up the sign? Wait, the inner shell has charge per unit length \( \lambda = -7.0\times 10^{-6} \, \text{C/m} \), so the enclosed charge per unit length is \( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \). The electric field formula for cylindrical symmetry is \( E = \frac{\lambda_{enclosed}}{2\pi\epsilon_0 r} \). Let's calculate the actual value:

\( \frac{1}{2\pi\epsilon_0} = \frac{1}{2\times 3.1416\times 8.85\times 10^{-12}} \approx \frac{1}{5.568\times 10^{-11}} \approx 1.796\times 10^{10} \, \text{N·m}^2/\text{C}^2 \)

Then, \( E = \frac{\lambda_{enclosed}}{r} \times \frac{1}{2\pi\epsilon_0} \)

\( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \), \( r = 0.04 \, \text{m} \)

So \( \frac{\lambda_{enclosed}}{r} = \frac{-7.0\times 10^{-6}}{0.04} = -1.75\times 10^{-4} \, \text{C/m}^2 \)

Then, \( E = -1.75\times 10^{-4} \times 1.796\times 10^{10} \approx -3.143\times 10^{6} \, \text{N/C} \). The magnitude is \( 3.1\times 10^{6} \, \text{N/C} \), but wait, the options have -6.3×10⁶, 7.9×10⁷, 2.3×10⁶, 3.1×10⁶. Wait, maybe I made a mistake in the enclosed charge. Wait, the two shells are concentric, thin-walled. The inner shell has radius \( r_i = 3.0 \, \text{cm} \), outer \( r_o = 6.0 \, \text{cm} \). The point is at \( r = 4.0 \, \text{cm} \), which is between \( r_i \) and \( r_o \). So the enclosed charge per unit length is only the inner shell's charge per unit length, which is \( \lambda_{inner} = -7.0\times 10^{-6} \, \text{C/m} \). The outer shell's charge is at \( r_o = 6.0 \, \text{cm} \), so it's outside the Gaussian surface (since \( r = 4.0 \, \text{cm} < 6.0 \, \text{cm} \)), so it's not enclosed. Wait, but maybe the problem is that the two shells are charged, so the inner is \( \lambda_1 = -7.0\times 10^{-6} \, \text{C/m} \), outer is \( \lambda_2 = 5.0\times 10^{-6} \, \text{C/m} \). Wait, no, the problem says "the charge per unit length is \( -7.0\times 10^{-6} \, \text{C/m} \) on the inner shell and \( 5.0\times 10^{-6} \, \text{C/m} \) on the outer shell". So at \( r = 4.0 \, \text{cm} \) (between \( r_i = 3.0 \, \text{cm} \) and \( r_o = 6.0 \, \text{cm} \)), the enclosed charge per unit length is only the inner shell's, \( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \). Then the electric field is \( E = \frac{\lambda_{enclosed}}{2\pi\epsilon_0 r} \). Let's recalculate:

\( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \)

\( r = 0.04 \, \text{m} \)

\( 2\pi\epsilon_0 = 2\times 3.1416\times 8.85\times 10^{-12} \approx 5.568\times 10^{-11} \, \text{C}^2/\text{N·m}^2 \)

So \( E = \frac{-7.0\times 10^{-6}}{5.568\times 10^{-11} \times 0.04} \)

Calculate denominator: \( 5.568\times 10^{-11} \times 0.04 = 2.227\times 10^{-12} \)

Then \( E = \frac{-7.0\times 10^{-6}}{2.227\times 10^{-12}} \approx -3.143\times 10^{6} \, \text{N/C} \). The magnitude is \( 3.1\times 10^{6} \, \text{N/C} \), but the options have -6.3×10⁶, which is double. Wait, maybe I misread the problem. Wait, maybe the two shells are both enclosed? No, \( r = 4.0 \, \text{cm} \) is between \( 3.0 \, \text{cm} \) and \( 6.0 \, \text{cm} \), so inner shell is inside (radius 3.0 cm), so the Gaussian surface at r=4.0 cm encloses the inner shell (radius 3.0 cm < 4.0 cm) but not the outer shell (radius 6.0 cm > 4.0 cm). Wait, unless the inner shell is the outer one? No, the problem says "inner shell" and "outer shell" with \( r_i = 3.0 \, \text{cm} \) and \( r_o = 6.0 \, \text{cm} \). So inner is smaller radius. So at r=4.0 cm, inside the outer shell, outside the inner shell? Wait, no: inner shell has radius 3.0 cm, so its surface is at 3.0 cm. So a point at 4.0 cm is outside the inner shell (since 4.0 > 3.0) and inside the outer shell (since 4.0 < 6.0). So the Gaussian surface at r=4.0 cm encloses the inner shell's charge. The outer shell's charge is at r=6.0 cm, so it's outside the Gaussian surface, so not enclosed. Wait, but maybe the problem is that the two shells are concentric, so the inner shell is inside the outer shell. So the charge per unit length on inner is \( \lambda_1 = -7.0\times 10^{-6} \, \text{C/m} \), outer is \( \lambda_2 = 5.0\times 10^{-6} \, \text{C/m} \). Then, at r=4.0 cm (between 3.0 and 6.0), the enclosed charge per unit length is \( \lambda_1 \), because the outer shell is at 6.0, so not enclosed. So the electric field should be due to \( \lambda_1 \). But the calculation gives ~ -3.1×10⁶ N/C, which is one of the options (3.1×10⁶ N/C, but with a negative sign? Wait, the first option is -6.3×10⁶, which is double. Wait, maybe I made a mistake in the formula. Wait, the formula for electric field due to a long charged cylinder (or cylindrical shell) is \( E = \frac{\lambda}{2\pi\epsilon_0 r} \) for \( r > R \) (where R is the radius of the cylinder). Wait, in this case, the inner shell has radius 3.0 cm, so at r=4.0 cm (which is > 3.0 cm), so we use the formula for outside the inner shell. So the electric field is \( E = \frac{\lambda_{inner}}{2\pi\epsilon_0 r} \). But maybe the problem is that the two shells are both charged, and we need to consider the net charge? Wait, no, the outer shell is at 6.0 cm, so at r=4.0 cm, the outer shell's charge is not enclosed. Wait, unless the inner shell is the outer one? No, the problem says inner and outer with \( r_i = 3.0 \) and \( r_o = 6.0 \). So inner is smaller. So at r=4.0, inside outer, outside inner. So enclosed charge is inner's charge. Wait, but let's recalculate the value:

\( \lambda = -7.0\times 10^{-6} \, \text{C/m} \)

\( r = 0.04 \, \text{m} \)

\( \frac{1}{2\pi\epsilon_0} = 1.796\times 10^{10} \, \text{N·m}^2/\text{C}^2 \) (as before)

So \( E = \frac{\lambda}{r} \times \frac{1}{2\pi\epsilon_0} = \frac{-7.0\times 10^{-6}}{0.04} \times 1.796\times 10^{10} \)

\( \frac{-7.0\times 10^{-6}}{0.04} = -1.75\times 10^{-4} \, \text{C/m}^2 \)

\( -1.75\times 10^{-4} \times 1.796\times 10^{10} = -3.143\times 10^{6} \, \text{N/C} \)

The magnitude is \( 3.1\times 10^{6} \, \text{N/C} \), which is one of the options (3.1×10⁶ N/C). Wait, but the first option is -6.3×10⁶, which is double. Maybe I misread the charge per unit length. Wait, the problem says "the charge per unit length is \( -7.0\times 10^{-6} \, \text{C/m} \) on the inner shell and \( 5.0\times 10^{-6} \, \text{C/m} \) on the outer shell". Wait, maybe the enclosed charge is the inner plus the outer? No, because the outer shell is at r=6.0 cm, so at r=4.0 cm, the outer shell is outside the Gaussian surface, so its charge is not enclosed. Only the inner shell's charge is enclosed. So the enclosed charge per unit length is \( \lambda_{enclosed} = -7.0\times 10^{-6} \, \text{C/m} \). Then the electric field is as calculated. So the answer should be 3.1×10⁶ N/C, which is one of the options. Wait, but let's check the calculation again. Wait, maybe the formula is \( E = \frac{\lambda_{enclosed}}{2\pi\epsilon_0 r} \), and \( \lambda_{enclosed} \) is the net charge per unit length enclosed. So if the inner shell has \( \lambda_1 = -7.0\times 10^{-6} \) and the outer shell has \( \lambda_2 = 5.0\times 10^{-6} \), but at r=4.0 cm, only \( \lambda_1 \) is enclosed, so \( \lambda_{enclosed} = -7.0\times 10^{-6} \). Then the calculation is correct. So the magnitude is approximately 3.1×10⁶ N/C, which is option 3.1×10⁶ N/C.

Answer:

3.1×10⁶ N/C (corresponding to the option "3.1×10⁶ N/C")