Physics

Electric Field and Force Analysis of an Electric Dipole System

• Dipole Configuration Setup:

  • Two source charges of equal magnitude $Q$ and opposite signs are positioned symmetrically along the x-axis at distance $a$ from the origin:

  • Charge $+Q$ is located at position $(a, 0)$.

  • Charge $-Q$ is located at position $(-a, 0)$.

  • Target evaluation point $P$ is located on the y-axis at height $b$, with coordinates $(0, b)$.

• Displacement Vectors to Point P:

  • Displacement vector $\bar{d}_1$ from source charge $+Q(a, 0)$ to point $P(0, b)$:     $\bar{d}_1 = (0 - a, b - 0) = (-a, b)$

  • Displacement vector $\bar{d}_2$ from source charge $-Q(-a, 0)$ to point $P(0, b)$:     $\bar{d}_2 = (0 - (-a), b - 0) = (a, b)$

• Distances and Unit Vectors:

  • By the Pythagorean theorem, distance $d_1$ and distance $d_2$ are equal:     $d_1 = d_2 = \sqrt{a^2 + b^2}$

  • Squared distance terms:     $d_1^2 = d_2^2 = a^2 + b^2$

  • Unit direction vector $\hat{d}_1$ pointing from $+Q$ to $P$:     $\hat{d}_1 = \frac{(-a, b)}{\sqrt{a^2 + b^2}}$

  • Unit direction vector $\hat{d}_2$ pointing from $-Q$ to $P$:     $\hat{d}_2 = \frac{(a, b)}{\sqrt{a^2 + b^2}}$

• Coulombic Force on a Test Charge:

  • Place a test charge $q$ at point $P(0, b)$.

  • By the principle of superposition, total Coulombic force $\bar{F}$ on test charge $q$ is:     $\bar{F} = \bar{F}_{+Q} + \bar{F}_{-Q}$     $\bar{F} = K_e \frac{Q q}{d_1^2} \hat{d}_1 + K_e \frac{(-Q) q}{d_2^2} \hat{d}_2$

  • Substituting distances and unit vectors:     $\bar{F} = K_e \frac{Q q}{a^2 + b^2} \left( \frac{(-a, b)}{\sqrt{a^2 + b^2}} \right) + K_e \frac{(-Q) q}{a^2 + b^2} \left( \frac{(a, b)}{\sqrt{a^2 + b^2}} \right)$

  • Factoring out common numerical terms and combining fractional powers using exponent rules ($n^1 \times n^{1/2} = n^{3/2}$):     $\bar{F} = \frac{K_e Q q}{(a^2 + b^2)^{3/2}} (-a, b) - \frac{K_e Q q}{(a^2 + b^2)^{3/2}} (a, b)$

  • Vector arithmetic evaluation:

  • x-component: $-a - a = -2a$

  • y-component: $b - b = 0$     $\bar{F} = q \times \frac{K_e Q}{(a^2 + b^2)^{3/2}} (-2a, 0)$

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