A parallel plate capacitor has a capacitance of $50 ext{ pF}$ and is charged to a potential difference of $100 ext{ V}$ by a battery. The battery is then disconnected, and a dielectric slab of dielectric constant $K = 4$ (which completely fills the space between the plates) is inserted. Find the new capacitance of the capacitor. Find the new potential difference across the plates. Find the change in the electrostatic energy stored in the capacitor.

Q: A parallel plate capacitor has a capacitance of $50\text{ pF}$ and is charged to a potential difference of $100\text{ V}$ by a battery. The battery is then disconnected, and a dielectric slab of dielectric constant $K = 4$ (which completely fills the space between the plates) is inserted. Find the new capacitance of the capacitor. Find the new potential difference across the plates. Find the change in the electrostatic energy stored in the capacitor.

Given Data: Initial capacitance ($C0$) = $50\text{ pF} = 50 \times 10^{-12}\text{ F}$ Initial potential difference ($V0$) = $100\text{ V}$ Dielectric constant ($K$) = $4$ Since the battery is disconnected before inserting the dielectric slab, the charge on the capacitor remains constant ($Q = Q0$). --- Step 1: Find the new capacitance ($C$)\nWhen a dielectric slab of constant $K$ is introduced, the new capacitance becomes $K$ times the initial capacitance. $$C = K \times C0$$ $$C = 4 \times 50\text{ pF} = 200\text{ pF} = 2.0 \times 10^{-10}\text{ F}$$ Answer 1: The new capacitance is $200\text{ pF}$. --- Step 2: Find the new potential difference ($V$)\nThe charge $Q$ on the capacitor before and after inserting the dielectric is constant.\nInitial charge $Q0 = C0 \times V0$ $$Q0 = (50 \times 10^{-12}\text{ F}) \times (100\text{ V}) = 5 \times 10^{-9}\text{ C}$$\nSince $Q = Q0$, the new potential difference $V$ is given by: $$V = \frac{Q}{C} = \frac{Q0}{K \cdot C0} = \frac{V0}{K}$| $$V = \frac{100\text{ V}}{4} = 25\text{ V}$$ Answer 2: The new potential difference across the plates is $25\text{ V}$. --- Step 3: Find the change in electrostatic energy ($\Delta U$)\nInitial electrostatic energy ($U0$): $$U0 = \frac{1}{2} C0 V0^2$$ $$U0 = \frac{1}{2} \times (50 \times 10^{-12}\text{ F}) \times (100\text{ V})^2$$ $$U0 = \frac{1}{2} \times 50 \times 10^{-12} \times 10000 = 2.5 \times 10^{-7}\text{ J} = 250\text{ nJ}$| \nFinal electrostatic energy ($U$): $$U = \frac{1}{2} C V^2 \quad \text{or} \quad U = \frac{Q0^2}{2C} = \frac{U0}{K}$$ $$U = \frac{2.5 \times 10^{-7}\text{ J}}{4} = 0.625 \times 10^{-7}\text{ J} = 62.5\text{ nJ}$| \nChange in energy (decrease in energy): $$\Delta U = U - U0 = 62.5\text{ nJ} - 250\text{ nJ} = -187.5\text{ nJ}$| Answer 3: The electrostatic energy stored in the capacitor decreases by $1.875 \times 10^{-7}\text{ J}$ (or $187.5\text{ nJ}$).

Question

A parallel plate capacitor has a capacitance of $50	ext{ pF}$ and is charged to a potential difference of $100	ext{ V}$ by a battery. The battery is then disconnected, and a dielectric slab of dielectric constant $K = 4$ (which completely fills the space between the plates) is inserted. Find the new capacitance of the capacitor. Find the new potential difference across the plates. Find the change in the electrostatic energy stored in the capacitor.

Accepted Answer

Given Data: Initial capacitance ($C0$) = $50	ext{ pF} = 50 	imes 10^{-12}	ext{ F}$ Initial potential difference ($V0$) = $100	ext{ V}$ Dielectric constant ($K$) = $4$ Since the battery is disconnected before inserting the dielectric slab, the charge on the capacitor remains constant ($Q = Q0$). --- Step 1: Find the new capacitance ($C$)
When a dielectric slab of constant $K$ is introduced, the new capacitance becomes $K$ times the initial capacitance. $$C = K 	imes C0$$ $$C = 4 	imes 50	ext{ pF} = 200	ext{ pF} = 2.0 	imes 10^{-10}	ext{ F}$$ Answer 1: The new capacitance is $200	ext{ pF}$. --- Step 2: Find the new potential difference ($V$)
The charge $Q$ on the capacitor before and after inserting the dielectric is constant.
Initial charge $Q0 = C0 	imes V0$ $$Q0 = (50 	imes 10^{-12}	ext{ F}) 	imes (100	ext{ V}) = 5 	imes 10^{-9}	ext{ C}$$
Since $Q = Q0$, the new potential difference $V$ is given by: $$V = \frac{Q}{C} = \frac{Q0}{K \cdot C0} = \frac{V0}{K}$| $$V = \frac{100	ext{ V}}{4} = 25	ext{ V}$$ Answer 2: The new potential difference across the plates is $25	ext{ V}$. --- Step 3: Find the change in electrostatic energy ($\Delta U$)
Initial electrostatic energy ($U0$): $$U0 = \frac{1}{2} C0 V0^2$$ $$U0 = \frac{1}{2} 	imes (50 	imes 10^{-12}	ext{ F}) 	imes (100	ext{ V})^2$$ $$U0 = \frac{1}{2} 	imes 50 	imes 10^{-12} 	imes 10000 = 2.5 	imes 10^{-7}	ext{ J} = 250	ext{ nJ}$| 
Final electrostatic energy ($U$): $$U = \frac{1}{2} C V^2 \quad 	ext{or} \quad U = \frac{Q0^2}{2C} = \frac{U0}{K}$$ $$U = \frac{2.5 	imes 10^{-7}	ext{ J}}{4} = 0.625 	imes 10^{-7}	ext{ J} = 62.5	ext{ nJ}$| 
Change in energy (decrease in energy): $$\Delta U = U - U0 = 62.5	ext{ nJ} - 250	ext{ nJ} = -187.5	ext{ nJ}$| Answer 3: The electrostatic energy stored in the capacitor decreases by $1.875 	imes 10^{-7}	ext{ J}$ (or $187.5	ext{ nJ}$).

Given Data:

Step 1: Find the new capacitance ($C$)\nWhen a dielectric slab of constant $K$ is introduced, the new capacitance becomes $K$ times the initial capacitance.

Step 2: Find the new potential difference ($V$)\nThe charge $Q$ on the capacitor before and after inserting the dielectric is constant.\nInitial charge $Q_0 = C_0 \times V_0$

Step 3: Find the change in electrostatic energy ($\Delta U$)\nInitial electrostatic energy ($U_0$):