Question

23.6 applying gauss law: spherical symmetry two charged concentric spherical shells have radii 10.0 cm and 15.0 cm. the charge on the inner shell is 10.00 × 10⁻⁶ c, and that on the outer shell is 4.00 × 10⁻⁶ c. by what factor will the electric field at r = 30.0 cm change if the inner shell’s radius is halved? options: 2, 1, 1/2, 0 save for later submit answer

Explanation:

Step1: Recall Gauss's Law for Electric Field

For a spherical charge distribution, the electric field at a distance \( r \) (where \( r \) is greater than the radius of the outermost charged shell) is given by \( E = \frac{kQ_{\text{total}}}{r^2} \), where \( k=\frac{1}{4\pi\epsilon_0} \) and \( Q_{\text{total}} \) is the total charge enclosed by the Gaussian surface of radius \( r \).

Here, the two concentric spherical shells have charges \( Q_{\text{inner}} = 10.00\times 10^{-6}\, \text{C} \) and \( Q_{\text{outer}} = 4.00\times 10^{-6}\, \text{C} \). The point where we measure the electric field is \( r = 30.0\, \text{cm} \), which is greater than the radius of the outer shell (\( 15.0\, \text{cm} \)). So the total charge enclosed \( Q_{\text{enclosed}}=Q_{\text{inner}} + Q_{\text{outer}} \), and the electric field at \( r = 30.0\, \text{cm} \) depends on \( Q_{\text{enclosed}} \) and \( r \).

Step2: Analyze the effect of halving the inner shell's radius

The radius of the inner shell is initially \( r_{\text{inner, initial}} = 10.0\, \text{cm} \), and after halving, \( r_{\text{inner, final}}=\frac{10.0}{2} = 5.0\, \text{cm} \). However, the charge on the inner shell (\( Q_{\text{inner}} \)) and the charge on the outer shell (\( Q_{\text{outer}} \)) remain the same. Also, the distance \( r = 30.0\, \text{cm} \) (where we measure the electric field) is still greater than the radius of the outer shell (which is \( 15.0\, \text{cm} \), unchanged) and also greater than the new radius of the inner shell (\( 5.0\, \text{cm} \)).

The total charge enclosed by the Gaussian surface of radius \( r = 30.0\, \text{cm} \) is \( Q_{\text{enclosed}}=Q_{\text{inner}} + Q_{\text{outer}} \), which does not change because the charges on the shells are unchanged. The distance \( r \) from the center to the point where we measure the electric field is also unchanged.

From the formula \( E=\frac{kQ_{\text{enclosed}}}{r^2} \), since both \( Q_{\text{enclosed}} \) and \( r \) are unchanged, the electric field \( E \) will not change. So the factor by which the electric field changes is \( 1 \) (because the new electric field \( E_{\text{new}} \) is equal to the old electric field \( E_{\text{old}} \), so \( \frac{E_{\text{new}}}{E_{\text{old}}}=1 \)).

Answer:

\( 1 \)