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Step-by-Step Solution
Step 1: Identify the initial charge on the capacitor
The capacitor of capacitance $C = 90\,\text{pF}$ is connected to a $20\,\text{V}$ battery. Initially (before the dielectric is inserted), the charge on the capacitor is given by:
$Q_i = CV = (90 \times 10^{-12}\,\text{F}) \times (20\,\text{V}) = 1.8 \times 10^{-9}\,\text{C} = 1.8\,\text{nC}.$
Step 2: Determine the capacitance with the dielectric inserted
When the dielectric of dielectric constant $K = \frac{5}{3}$ is inserted between the plates, the new capacitance becomes:
$C_{\text{new}} = K \cdot C = \frac{5}{3} \times 90 \times 10^{-12}\,\text{F} = 150 \times 10^{-12}\,\text{F} = 150\,\text{pF}.$
Because the battery remains connected, the voltage across the capacitor is still $20\,\text{V}$.
Step 3: Calculate the final charge on the capacitor
With the dielectric in place, the final charge is:
$Q_f = C_{\text{new}} \times V = (150 \times 10^{-12}\,\text{F}) \times (20\,\text{V}) = 3.0 \times 10^{-9}\,\text{C} = 3.0\,\text{nC}.$
Step 4: Compute the induced charge
The induced charge on the dielectric is the difference between the final and the initial charge:
$\Delta Q_{\text{induced}} = Q_f - Q_i = 3.0\,\text{nC} - 1.8\,\text{nC} = 1.2\,\text{nC}.$
Step 5: Conclude the result
Hence, the magnitude of the induced charge is $1.2\,\text{nC}.$
