1. **State the problem:** We have a capacitor with plate separation $d$ and initial capacitance $C_0 = 1.7 \mu F$. A Teflon dielectric slab of thickness $\frac{3d}{4}$ and dielectric constant $K=2.1$ is inserted between the plates. We need to find the new capacitance $C$. 2. **Understand the setup:** The capacitor now consists of two capacitors in series: - One with dielectric thickness $\frac{3d}{4}$ and dielectric constant $K=2.1$. - One with air (dielectric constant $1$) thickness $d - \frac{3d}{4} = \frac{d}{4}$. 3. **Capacitance formulas:** For a parallel plate capacitor, capacitance is $C = \frac{\epsilon A}{d}$, where $\epsilon = K \epsilon_0$. 4. **Calculate individual capacitances:** - Capacitance with dielectric: $$C_1 = \frac{K \epsilon_0 A}{\frac{3d}{4}} = \frac{4K \epsilon_0 A}{3d}$$ - Capacitance with air: $$C_2 = \frac{\epsilon_0 A}{\frac{d}{4}} = \frac{4 \epsilon_0 A}{d}$$ 5. **Express $C_0$:** Initial capacitance without dielectric: $$C_0 = \frac{\epsilon_0 A}{d}$$ 6. **Rewrite $C_1$ and $C_2$ in terms of $C_0$:** $$C_1 = \frac{4K \epsilon_0 A}{3d} = \frac{4K}{3} C_0$$ $$C_2 = \frac{4 \epsilon_0 A}{d} = 4 C_0$$ 7. **Calculate total capacitance $C$ for series capacitors:** $$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{\frac{4K}{3} C_0} + \frac{1}{4 C_0} = \frac{3}{4K C_0} + \frac{1}{4 C_0} = \frac{3}{4K C_0} + \frac{1}{4 C_0}$$ 8. **Combine terms:** $$\frac{1}{C} = \frac{3}{4K C_0} + \frac{1}{4 C_0} = \frac{3 + K}{4 K C_0}$$ 9. **Invert to find $C$:** $$C = \frac{4 K C_0}{3 + K}$$ 10. **Substitute values:** $$K = 2.1, \quad C_0 = 1.7 \mu F$$ $$C = \frac{4 \times 2.1 \times 1.7}{3 + 2.1} = \frac{14.28}{5.1} \approx 2.8 \mu F$$ **Final answer:** $$\boxed{C \approx 2.8 \mu F}$$