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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Capacitance of the capacitor (without dielectric): C = 200 \times 10^{-6}\,\text{F} (200 μF)
• Potential difference across the capacitor: V = 200\,\text{V}
• Dielectric constant of inserted slab: K = 2
• The battery remains connected, so the voltage remains constant at 200 V.
Step 2: Write the Formula for the Change in Electrostatic Energy
When a dielectric slab with dielectric constant K is inserted into a capacitor of capacitance C (while connected to a constant voltage V ), the change in energy \Delta U is given by
\Delta U = \frac{1}{2} \bigl(\Delta C\bigr) V^2,
where \Delta C = (K - 1)C . Hence,
\Delta U = \frac{1}{2}(K - 1) C V^2.
Step 3: Compute the Change in Capacitance
The initial capacitance is C , and the final capacitance after inserting the dielectric is K C . Thus, the increase in capacitance is
\Delta C = K C - C = (K - 1)\,C.
In this problem, K = 2 and C = 200 \times 10^{-6}\,\text{F} , so
\Delta C = (2 - 1)\times 200\times 10^{-6}\,\text{F} = 200 \times 10^{-6}\,\text{F}.
Step 4: Substitute Values into the Formula
Substitute K = 2 , C = 200 \times 10^{-6}\,\text{F} , and V = 200\,\text{V} into the expression for \Delta U :
\Delta U = \frac{1}{2}(2 - 1)\times \bigl(200 \times 10^{-6}\,\text{F}\bigr) \times (200\,\text{V})^2.
Step 5: Numerical Calculation
Perform each step:
Compute (2 - 1) = 1 .
(200\,\text{V})^2 = 40000\,\text{V}^2 .
So,
\Delta U = \frac{1}{2} \times 1 \times 200 \times 10^{-6}\,\text{F} \times 40000\,\text{V}^2.
Combine the numerical factors:
\Delta U = \frac{1}{2} \times 200 \times 10^{-6} \times 40000.
First multiply 200 \times 40000 = 8 \times 10^6 , so
\Delta U = \frac{1}{2} \times (8 \times 10^6) \times 10^{-6}\,\text{J} = \frac{1}{2} \times 8\,\text{J} = 4\,\text{J}.
Final Answer
The change in the electrostatic energy in the capacitor is 4\,\text{J} .
