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Step-by-Step Solution
Step 1: Identify the known quantities
• Initial capacitance of the capacitor, $C = 12\,\text{pF} = 12 \times 10^{-12}\,\text{F}$.
• Potential difference before the dielectric is inserted, $V = 10\,\text{V}$.
• Dielectric constant of the porcelain slab, $k = 6.5$.
• Since the battery is disconnected, the charge on the capacitor remains constant.
Step 2: Calculate the initial charge on the capacitor
The charge on the capacitor is given by
$Q = C \times V.$
Substituting the values:
$Q = (12 \times 10^{-12}\,\text{F}) \times (10\,\text{V}) = 120 \times 10^{-12}\,\text{C} = 120\,\text{pC}.$
Step 3: Calculate the initial energy of the capacitor
Before the dielectric is inserted, the energy stored in the capacitor is
$U_i = \tfrac{1}{2} \, C \, V^2.$
Substituting the initial values:
$U_i = \tfrac{1}{2} \times \bigl(12 \times 10^{-12}\,\text{F}\bigr) \times (10\,\text{V})^2.$
Therefore,
$U_i = \tfrac{1}{2} \times 12 \times 10^{-12} \times 100 \,\text{J} = 600 \times 10^{-12}\,\text{J} = 600\,\text{pJ}.$
Step 4: Determine the new capacitance after inserting the dielectric
When a dielectric of constant $k$ is inserted fully between the plates (and the geometry remains the same), the new capacitance becomes
$C' = k \, C.$
Thus,
$C' = 6.5 \times (12 \times 10^{-12})\,\text{F} = 78 \times 10^{-12}\,\text{F} = 78\,\text{pF}.$
Step 5: Calculate the final voltage across the capacitor
Since the charge remains the same (battery is disconnected), the final voltage across the capacitor is
$V_f = \frac{Q}{C'}.$
Using $Q = 120\,\text{pC}$ and $C' = 78\,\text{pF}$:
$V_f = \frac{120 \times 10^{-12}\,\text{C}}{78 \times 10^{-12}\,\text{F}} = \frac{120}{78}\,\text{V} \approx 1.54\,\text{V}.$
Step 6: Compute the final energy of the capacitor
The final energy stored in the capacitor is
$U_f = \frac{Q^2}{2\,C'}.$
Substituting the values:
$U_f = \frac{(120 \times 10^{-12}\,\text{C})^2}{2 \times 78 \times 10^{-12}\,\text{F}}.$
Carrying out the arithmetic carefully:
$Q^2 = (120 \times 10^{-12})^2 = 14400 \times 10^{-24}\,\text{C}^2 = 1.44 \times 10^{-20}\,\text{C}^2,$
so
$U_f = \frac{1.44 \times 10^{-20}}{2 \times 78 \times 10^{-12}}\,\text{J} = \frac{1.44 \times 10^{-20}}{156 \times 10^{-12}}\,\text{J}.$
$U_f \approx \frac{1.44}{156} \times 10^{-8}\,\text{J} \approx 0.00923 \times 10^{-8}\,\text{J} = 92.3 \times 10^{-12}\,\text{J} = 92.3\,\text{pJ}.$
Step 7: Calculate the work done by the capacitor on the dielectric slab
The initial energy was $U_i$, and the final energy is $U_f$. The decrease in the capacitor’s energy (which is the work done on the slab) is
$W = U_i - U_f.$
Thus,
$W = 600\,\text{pJ} - 92.3\,\text{pJ} \approx 507.7\,\text{pJ}.$
Rounded suitably, this is approximately $508\,\text{pJ}.$
Final Answer
$\boxed{508\,\text{pJ}}$
