Think of the force on the top-right charge q as a sum of forces from the other three charges (superposition). The two charges at the adjacent corners are at distance equal to the side of the square, so the forces from them on the top-right charge have equal magnitudes and point left and down, respectively. Specifically, each of those forces has magnitude F0 = k q (10 nC) / s^2, with one directed to the left and the other downward. The third charge, bottom-left, is diagonally opposite and sits a distance √2 s away. The force it exerts on the top-right charge has magnitude Fdiag = k q |Q| / (√2 s)^2 = k q |Q| / (2 s^2). Its direction depends on the sign of Q: if Q is positive, the force is along the diagonal away from bottom-left, i.e., up and to the right. For the net force on the top-right charge to be zero, the diagonal force must supply horizontal and vertical components that exactly cancel the leftward and downward forces. Along the diagonal (45 degrees), each component is Fdiag/√2. Thus we need: Fdiag/√2 = F0. Plugging in F0 = k q (10 nC) / s^2 and Fdiag = k q |Q| / (2 s^2), (k q |Q| / (2 s^2)) / √2 = k q (10 nC) / s^2. Cancel common factors to get |Q|/2 = √2 × 10 nC, so |Q| = 20 √2 nC ≈ 28 nC. Sign must be positive to point the diagonal force up and to the right, which cancels the left and down forces. Therefore Q ≈ +28 nC.

At the center of a circular loop, the magnetic field from a current I is B = μ0 I / (2R). This shows the field is inversely proportional to the radius for a fixed current. If the radius is doubled, the denominator doubles, so the field becomes half of its original value. The intuition is that spreading the same current over a larger loop weakens the magnetic influence at the center. The magnitude is halved (the direction would still be along the axis given by the right-hand rule, if you care about orientation).

Brewster’s angle is the incidence angle at which the reflected light becomes perfectly polarized parallel to the interface. At this angle, the reflected and refracted rays are perpendicular to each other, which gives a simple relationship from Snell’s law. From Snell’s law, n1 sin θi = n2 sin θt, and at Brewster’s angle θi + θt = 90°, so sin θt = cos θi. This leads to n1 sin θi = n2 cos θi, and rearranging gives tan θi = n2 / n1. Plugging in n1 = 1.00 (air) and n2 ≈ 1.50 (glass), tan θB = 1.50, so θB = arctan(1.50) ≈ 56.3°. Therefore the Brewster angle for this air-to-glass interface is about 56.3°. The other angles don’t satisfy tan θ = 1.50 (for example, tan 34° ≈ 0.674, tan 45° = 1, tan 30° ≈ 0.577), so they aren’t Brewster angles here.

Ampere's law applied to an ideal long solenoid gives a uniform magnetic field inside and negligible field outside. If you draw a rectangular Amperian loop with a long side of length L along the solenoid’s axis, the line integral is essentially B times L inside (the outside contribution is zero). The current enclosed by the loop is the total current from all turns within length L, which is (turns per unit length) times L times the current per turn: N L I. Ampere’s law then reads B L = μ0 (N L I), so the magnetic field inside is B = μ0 N I. This shows the field grows linearly with both the current and the density of turns.