A parallel plate capacitor consisting of two circular plates of radius 10mathrm~cm$10\mathrm{~cm}$ is being charged by a constant current of 0.15mathrm~A$0.15\mathrm{~A}$. If the rate of change of potential difference between the plates is 7 times 10^8mathrm~V/s$7 \times 10^{8}\mathrm{~V/s}$ then the integer value of the distance between the parallel plates is ______ mumathrmm$\mu\mathrm{m}$. left(textTake epsilon_0 = 9 times 10^-12fracmathrmFmathrmm, pi = frac227right)$$\left(\text{Take } \epsilon_{0} = 9 \times 10^{-12}\frac{\mathrm{F}}{\mathrm{m}}, \pi = \frac{22}{7}\right)$$$$$$

### Related Formula rho_E = frac12varepsilon_0 E_rms^2$$\rho_E = \frac{1}{2}\varepsilon_0 E_{rms}^2$$ rho_B = fracB_rms^22mu_0$$\rho_B = \frac{B_{rms}^2}{2\mu_0}$$ p_r = fracIc$$p_r = \frac{I}{c}$$ (radiation pressure for completely absorbing surface) ### Core Logic Statement I: In an EM wave, energy is stored in both the electric and magnetic fields. The energy densities are equal: U_E = U_B$U_E = U_B$ because E = cB$E = cB$ and c^2 = frac1mu_0 epsilon_0$c^2 = \frac{1}{\mu_0 \epsilon_0}$. This is a correct fact. Statement II: Electromagnetic waves carry momentum (p = U/c$p = U/c$). When they strike a surface, they transfer this momentum, creating radiation pressure. This is also a correct fact. ### Step 1: Final Evaluation Since both facts are universally true theoretical properties of EM waves, both statements are correct. ### Pattern Recognition Standard NCERT theory. Energy density is strictly symmetric 50-50 between E and B. Momentum transfer to$\to$ pressure is the definitive signature of the particle-like nature (photons) of EM waves. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula U_texttotal average = frac12epsilon_0 E_0^2$$U_{\text{total average}} = \frac{1}{2}\epsilon_0 E_0^2$$ ### Core Logic For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: U_textavg = U_E + U_B = 2U_E = 2 left( frac14epsilon_0 E_0^2 right) = frac12epsilon_0 E_0^2$$U_{\text{avg}} = U_E + U_B = 2U_E = 2 \left( \frac{1}{4}\epsilon_0 E_0^2 \right) = \frac{1}{2}\epsilon_0 E_0^2$$ Where E_0$E_0$ is the amplitude of the electric field. ### Step 2: Substitution Given: E_0 = 50 mathrm\, V/m$E_0 = 50 \mathrm{\, V/m}$ epsilon_0 = 8.85 times 10^-12 mathrm\, C^2/(Ncdot m^2)$\epsilon_0 = 8.85 \times 10^{-12} \mathrm{\, C^2/(N\cdot m^2)}$ U_textavg = frac12 times (8.85 times 10^-12) times (50)^2$$U_{\text{avg}} = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (50)^2$$ U_textavg = frac12 times 8.85 times 10^-12 times 2500$$U_{\text{avg}} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500$$ U_textavg = 1.10625 times 10^-8 mathrm\, J/m^3$$U_{\text{avg}} = 1.10625 \times 10^{-8} \mathrm{\, J/m^3}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves