Let's break down the problem step by step: ### **Given Data** - Two point charges: \( +q \) and \( -q \) separated by distance \( d \). - Both are aligned parallel with the y-axis and are placed symmetrically about the x-axis at positions: - \( +q \) at \( (0, +d/2) \) - \( -q \) at \( (0, -d/2) \) - The Coulomb force constant, \( k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). - Dipole moment: \( p = q \cdot d \). - Point of interest: On x-axis at \( x \gg d \). --- ## **Step 1: Expression for Potential \( V \) on the x-Axis** The electric potential at a point due to a point charge is: \[ V = k \frac{q}{r} \] For the two charges: - Distance from \( +q \) to point \( x \) on x-axis: \( r_+ = \sqrt{x^2 + (d/2)^2} \) - Distance from \( -q \) to point \( x \): \( r_- = \sqrt{x^2 + (d/2)^2} \) The potential at point \( x \) due to both charges: \[ V(x) = k \left[ \frac{q}{\sqrt{x^2 + (d/2)^2}} - \frac{q}{\sqrt{x^2 + (d/2)^2}} \right] \] Because the distances are symmetric, but the charges are opposite, simplify: \[ V(x) = kq \left( \frac{1}{\sqrt{x^2 + (d/2)^2}} - \frac{1}{\sqrt{x^2 + (d/2)^2}} \right) \] But this is zero if we just plug in. **However**, for a dipole, we need to expand for \( x \gg d \). --- ## **Step 2: Dipole Approximation (\( x \gg d \))** For a dipole, the potential at a point on the axis (x-axis) is: \[ V(x) \approx \frac{k p}{x^2} \] where \( p = qd \). --- ## **Step 3: Plug in the Numbers** Given: - \( p = 5.98 \times 10^{-9} \) C·m - \( x = 37.1 \) cm \( = 0.371 \) m Plug into the formula: \[ V(x) = \frac{k p}{x^2} \] \[ V(x) = \frac{8.99 \times 10^9 \times 5.98 \times 10^{-9}}{(0.371)^2} \] Calculate numerator: \[ 8.99 \times 10^9 \times 5.98 \times 10^{-9} = 53.7602 \] Calculate denominator: \[ (0.371)^2 = 0.137641 \] So, \[ V(x) = \frac{53.7602}{0.137641} \approx 390.6 \, \text{V} \] --- ## **Step 4: Final Answer (with units and significant figures)** \[ \boxed{V = 391\, \text{V} } \] (Rounded to three significant figures.) --- ### **Summary** - **Expression for potential on x-axis (for \( x \gg d \)):** \[ V(x) = \frac{k p}{x^2} \] - **Potential at \( x = 37.1 \) cm:** \[ V = 391\, \text{V} \] --- Let me know if you need more detailed explanation or further clarifications!