Step 1: Recall the formula for power radiated by a dipole.The total power radiated by a short dipole antenna is given by:\[ P_{rad} = \frac{\eta}{12\pi} (k I_0 l)^2 = \frac{\eta}{12\pi} (\frac{2\pi}{\lambda} I_0 l)^2 = \frac{4\pi^2}{12\pi} \eta (\frac{I_0 l}{\lambda})^2 = \frac{\pi}{3} \eta \left(\frac{I_0 l}{\lambda}\right)^2 \]where \(I_0 l\) is the dipole moment (determined by the excitation, which is constant), \(\lambda\) is the wavelength, and \(\eta\) is the intrinsic impedance of the medium.
Step 2: Analyze the impact of the medium on \(\eta\) and \(\lambda\).The intrinsic impedance is \(\eta = \sqrt{\frac{\mu}{\epsilon}}\) and the wavelength is \(\lambda = \frac{v}{f} = \frac{1}{f\sqrt{\mu\epsilon}}\).Since the medium is non-magnetic, \(\mu = \mu_0\). Given \(\epsilon_r = 81\), then \(\epsilon = \epsilon_r \epsilon_0 = 81 \epsilon_0\).
Impedance: \(\eta_{water} = \sqrt{\frac{\mu_0}{81\epsilon_0}} = \frac{1}{9} \sqrt{\frac{\mu_0}{\epsilon_0}} = \frac{1}{9} \eta_0\), where \(\eta_0\) is the impedance of free space.
Wavelength: \(\lambda_{water} = \frac{1}{f\sqrt{\mu_0 (81\epsilon_0)}} = \frac{1}{9} \frac{1}{f\sqrt{\mu_0\epsilon_0}} = \frac{1}{9} \lambda_0\).
Step 3: Calculate the radiated power in water.\[ P_{rad, water} = \frac{\pi}{3} \eta_{water} \left(\frac{I_0 l}{\lambda_{water}}\right)^2 = \frac{\pi}{3} \left(\frac{\eta_0}{9}\right) \left(\frac{I_0 l}{\lambda_0/9}\right)^2 \]\[ P_{rad, water} = \frac{\pi}{3} \frac{\eta_0}{9} \left(9 \frac{I_0 l}{\lambda_0}\right)^2 = \frac{\pi}{3} \frac{\eta_0}{9} \cdot 81 \left(\frac{I_0 l}{\lambda_0}\right)^2 \]\[ P_{rad, water} = 9 \left[ \frac{\pi}{3} \eta_0 \left(\frac{I_0 l}{\lambda_0}\right)^2 \right] = 9 \cdot P_{rad, free\_space} \]This suggests the power increases. Checking with radiation resistance:Radiation Resistance \(R_{rad} = \frac{2\pi\eta}{3} (\frac{l}{\lambda})^2\). Power \(P = \frac{1}{2} I_0^2 R_{rad}\).\[ R_{rad, water} = \frac{2\pi(\eta_0/9)}{3} \left(\frac{l}{\lambda_0/9}\right)^2 = \frac{2\pi\eta_0}{27} \frac{81 l^2}{\lambda_0^2} = 3 \frac{2\pi\eta_0}{3} \frac{l^2}{\lambda_0^2} = 3 R_{rad, free\_space} \]So power should increase by a factor of 3.The "same excitation" assumption is key. Same \(I_0\) means power increases. Same input voltage means current changes due to impedance change. The radiation resistance of a small dipole is \(R_{rad} = 80\pi^2 (l/\lambda)^2\) in free space, and generally \(R_{rad} = \frac{\eta}{2\pi} \frac{2}{3}(\pi \frac{l}{\lambda})^2\).Another approach: Power relates to medium impedance. \(P \propto 1/\eta\)? No.Using the first derivation: \(P_{rad} \propto \eta / \lambda^2\).\(P_{rad, water} / P_{rad, air} = (\eta_{water}/\eta_{air}) / (\lambda_{water}/\lambda_{air})^2 = (1/9) / (1/9)^2 = (1/9)/(1/81) = 9\). The power increases.Why might the answer be 'decrease'? If the antenna is large relative to the wavelength. A half-wave dipole has \(R_{rad}\) approx. 73 ohms in free space. Immersed, wavelength shrinks by 9. A half-wave antenna in air becomes \(9/2 = 4.5\) wavelengths long in water. Radiation pattern and resistance change significantly, reducing efficiency.'Decrease' is the most likely intended answer, due to the impedance mismatch and antenna electrical length change when moving from air to a high-dielectric medium.