answer:Given: - Surface charge density of the infinite surface: ( sigma = rho A ) (where ( A ) is the area) - Radius of the circular aperture: ( R ) - Distance from the center of the aperture: ( a ) - Permittivity of free space: ( epsilon_0 ) Consider an infinite plane sheet with a uniform charge density ( sigma ), lying in the xy plane. The electric field ( E ) above or below the sheet, a distance ( a ) away, is given by: [ E = frac{sigma}{2 epsilon_0} ] Now, let's analyze the effect of the circular aperture of radius ( R ). The z-axis passes through the center of the disk, normal to the plane of the sheet. We need to find the electric field at a distance ( a ) above the center on the z-axis. The electric field due to the infinite sheet without the aperture is: [ E_{sheet} = frac{sigma}{2 epsilon_0} ] The electric field ( E_1 ) due to the circular disk of radius ( R ) at a distance ( a ) above its center is: [ E_1 = frac{sigma}{2 epsilon_0} times left[ 1 - frac{a}{sqrt{a^2 + R^2}} right] ] The electric field at point ( a ) due to the remaining charged region is the difference between the electric field of the infinite sheet and the disk: [ E_{aperture} = E_{sheet} - E_1 ] [ E_{aperture} = frac{sigma}{2 epsilon_0} left( 1 - left[ 1 - frac{a}{sqrt{a^2 + R^2}} right] right) ] [ E_{aperture} = frac{sigma a}{2 epsilon_0 sqrt{a^2 + R^2}} ] So, the magnitude of the electric field at a height ( a ) above the center of the circular aperture is: [ E_{aperture} = frac{sigma a}{2 epsilon_0 sqrt{a^2 + R^2}} ]

answer:Sorting a partially sorted array with the given property does not guarantee a runtime of O(klog{k}). Even if subarrays of size k are individually sorted, the array as a whole may still require additional sorting. For instance, consider an example with k = 3: [1, 2, 10000, 3, 4, 10001, 10002, 10003, 10004]. In this case, the subarrays of size 3 are sorted, but the array is not fully sorted. To sort such an array, you would need to merge k sorted sequences, each of length approximately n/k. This operation can be done in O(n log k) time. Therefore, the big O runtime for sorting this partially sorted array is O(n log k).

answer:Let x be the distance moved by the canoe. Then, the distance moved by the child is (6 - x) m. The center of mass of the system remains stationary. Therefore, the center of mass of the child-canoe system before and after the child's movement should be the same. Initial center of mass: ``` X_initial = (m_child * 0 + m_canoe * 3) / (m_child + m_canoe) ``` Final center of mass: ``` X_final = (m_child * (6 - x) + m_canoe * (3 + x)) / (m_child + m_canoe) ``` Equating X_initial and X_final, we get: ``` x = 2 m ``` Therefore, the distance moved by the canoe is 2 m, and the distance moved by the child is 4 m. The ratio of the distance moved by the canoe to the distance moved by the child is: ``` 2 m / 4 m = 1:2 ```

answer:For the cylindrical capacitor with the following specifications: - Inner conductor radius (R_i) = 2.1 mm - Outer conductor radius (R_o) = 3.4 mm - Length (L) = 3.0 m - Potential difference (V) = 330 mV Part A: Calculate the capacitance per unit length (C/L): [ frac{C}{L} = frac{2pi epsilon_0}{lnleft(frac{R_o}{R_i}right)} ] [ frac{C}{L} = frac{2pi times 8.85 times 10^{-12}}{lnleft(frac{3.4}{2.1}right)} approx 1.15 times 10^{-10} , text{F/m} ] Part B: Find the charges on the inner and outer conductors: Since the potential difference (V) is given, we can use the capacitance (C/L) to find the charge on the inner conductor (Q), which is positive: [ Q = C times V = (1.15 times 10^{-10} , text{F/m}) times (3.0 , text{m}) times (0.33 , text{V}) approx 0.11 , text{nC} ] The charge on the outer conductor is equal in magnitude but opposite in sign to the inner conductor, so it is: [ Q_{text{outer}} = -0.11 , text{nC} ] Thus, the charges on both conductors are +0.11 nC and -0.11 nC, respectively.