New RBI Grade B 2025 Notification will come Soon! Enroll Here

    Question

    Charges +q and –q are placed at (–d, 0) and (+d,

    0). A third charge Q is placed at (0, x). For what value of x will the force on Q be maximum?
    A x = d Correct Answer Incorrect Answer
    B x = 2d Correct Answer Incorrect Answer
    C x = √2 d Correct Answer Incorrect Answer
    D x = d/√2 Correct Answer Incorrect Answer

    Solution

    Given:

    • Charge +q at (-d, 0)
    • Charge -q at (+d, 0)
    • Charge Q at (0, x)
    Forces on Q: The force from +q on Q:
    • Distance: r₁ = √(d² + x²)
    • Force magnitude: F₁ = k|qQ|/(d² + x²)
    • Direction: along the line from (-d, 0) to (0, x)
    The force from -q on Q:
    • Distance: r₂ = √(d² + x²)
    • Force magnitude: F₂ = k|qQ|/(d² + x²)
    • Direction: along the line from (+d, 0) to (0, x)
    Components analysis: For the force from +q:
    • x-component: F₁ₓ = F₁ × (d/√(d² + x²)) = kqQd/(d² + x²)^(3/2)
    • y-component: F₁ᵧ = F₁ × (x/√(d² + x²)) = kqQx/(d² + x²)^(3/2)
    For the force from -q:
    • x-component: F₂ₓ = -F₂ × (d/√(d² + x²)) = -kqQd/(d² + x²)^(3/2)
    • y-component: F₂ᵧ = F₂ × (x/√(d² + x²)) = kqQx/(d² + x²)^(3/2)
    Net force: The x-components cancel out: F₁ₓ + F₂ₓ = 0 The y-components add: Fᵧ = F₁ᵧ + F₂ᵧ = 2kqQx/(d² + x²)^(3/2) So the total force magnitude is: F = |Fᵧ| = 2k|qQ||x|/(d² + x²)^(3/2) Finding the maximum: To maximize F, I need to maximize f(x) = |x|/(d² + x²)^(3/2) For x > 0, this becomes f(x) = x/(d² + x²)^(3/2) Taking the derivative: f'(x) = [(d² + x²)^(3/2) - x · (3/2)(d² + x²)^(1/2) · 2x] / (d² + x²)³ f'(x) = [(d² + x²)^(1/2)[(d² + x²) - 3x²]] / (d² + x²)³ f'(x) = (d² - 2x²) / (d² + x²)^(5/2) Setting f'(x) = 0: d² - 2x² = 0 x² = d²/2 x = ±d/√2 Verification:
    • For x < d/√2: f'(x) > 0 (increasing)
    • For x > d/√2: f'(x) < 0 (decreasing)
    Therefore, the force on Q is maximum when: x = ±d/√2 The maximum occurs at both positive and negative values due to symmetry

    Practice Next

    Relevant for Exams: