To find the value of the capacitor in a parallel resonant circuit with given half power frequencies, we can use the relationship between the resonant frequency, resistance, and capacitance.

Given Information
- Resistance (R) = 2 kΩ = 2000 Ω
- Half power frequencies (f1 and f2) = 86 kHz and 90 kHz

Calculating Resonant Frequency
- **Resonant Frequency (f0)**:
- The resonant frequency can be calculated as the average of the half power frequencies:
- \( f_0 = \frac{f_1 + f_2}{2} = \frac{86 \, kHz + 90 \, kHz}{2} = 88 \, kHz \)

Calculating Bandwidth
- **Bandwidth (Δf)**:
- The bandwidth of the circuit can be determined by:
- \( \Delta f = f_2 - f_1 = 90 \, kHz - 86 \, kHz = 4 \, kHz \)

Calculating Quality Factor (Q)
- **Quality Factor (Q)**:
- The Quality Factor can be calculated using:
- \( Q = \frac{f_0}{\Delta f} = \frac{88 \, kHz}{4 \, kHz} = 22 \)

Capacitance Calculation
- **Capacitance (C)**:
- The resonant frequency of a parallel resonant circuit is given by:
- \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
- Rearranging gives:
- \( C = \frac{1}{(2\pi f_0)^2 L} \)
- With \( L = \frac{R}{\omega_0 Q} \) (where \( \omega_0 = 2\pi f_0 \)):
- Substituting values:
- \( \omega_0 = 2\pi \cdot 88 \times 10^3 \, rad/s \)
- \( L = \frac{2000}{(2\pi \cdot 88 \times 10^3) \cdot 22} \)
- Finally, substitute \( L \) back to find \( C \).

Conclusion
Using these calculations, you can derive the exact value of the capacitor. The process involves evaluating the resonant frequency, bandwidth, and Quality Factor to find the inductance and ultimately the capacitance.