Step 1: The poles of the system are given by the eigenvalues of the matrix \( A \).
Step 2: The characteristic equation is found by solving: \[ \det(A - \lambda I) = 0 \]
Step 3: Substituting \( A = \begin{bmatrix} 0 & 3 \\ 3 & 0 \end{bmatrix} \): \[ \begin{vmatrix} 0 - \lambda & 3 \\ 3 & 0 - \lambda \end{vmatrix} = 0 \]
Step 4: Computing the determinant: \[ (-\lambda)(-\lambda) - (3 \times 3) = \lambda^2 - 9 = 0 \]
Step 5: Solving for \( \lambda \): \[ \lambda^2 = 9 \] \[ \lambda = \pm j3 \]
Step 6: The poles of the system are at \( s = \pm j3 \).
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \(\_\_\_\_\_\). (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: