Calculus Practice Test 2026 | Exam Prep Questions And Answers

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.