Comprehensive Study Notes: Electrostatic Potential Due to a Point Charge

Electrostatic Potential Due to a Point Charge

  • Definition of Setup:

    • A source charge, denoted as +Q+Q, is placed at a specific origin point.
    • The goal is to determine the electrostatic potential at a point PP located at a distance rr from the source charge +Q+Q.
    • The potential at a point is defined by the work done in bringing a unit positive test charge from infinity (\infty) to that point PP against the electrostatic forces.
  • Fundamental Relationship:

    • The relationship between the electric potential (VV) and the electric field (EE) over a distance (dxdx) is expressed by the integral formula:     V=rEdxV = -\int_{\infty}^{r} E \, dx

Mathematical Derivation and Calculus Process

  • Electric Field Intensity (EE):

    • At any intermediate distance xx from the source charge +Q+Q, the magnitude of the electric field (EE) is given by Coulomb's Law:     E=kQx2E = \frac{kQ}{x^2}
    • Here, kk is the electrostatic constant (where k=14πϵ0k = \frac{1}{4\pi\epsilon_0}).
  • Setting up the Integral:

    • Substituting the expression for the electric field into the potential formula:     V=rkQx2dxV = -\int_{\infty}^{r} \frac{kQ}{x^2} \, dx
    • Since kk and QQ are constants with respect to the distance xx, they are taken outside the integral:     V=kQrx2dxV = -kQ \int_{\infty}^{r} x^{-2} \, dx
  • Application of the Integration Power Rule:

    • The general rule for integration is stated as:     xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}
    • Applying this to the term x2x^{-2}, we find the antiderivative:     x2dx=x2+12+1=x11=1x\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x}
  • Evaluating the Definite Integral:

    • Applying the antiderivative back into the potential equation with limits from \infty to rr:     V=kQ[1x]rV = -kQ \left[ -\frac{1}{x} \right]_{\infty}^{r}
    • The two negative signs (one from the formula and one from the antiderivative) cancel each other out, resulting in a positive expression:     V=kQ[1x]rV = kQ \left[ \frac{1}{x} \right]_{\infty}^{r}
    • Evaluating the limits (Upper Limit minus Lower Limit):     V=kQ(1r1)V = kQ \left( \frac{1}{r} - \frac{1}{\infty} \right)
    • Since the value of 1\frac{1}{\infty} is mathematically defined as 00, the expression simplifies significantly:     V=kQ(1r0)V = kQ \left( \frac{1}{r} - 0 \right)

Final Expression and Analytical Observations

  • The Potential Formula:

    • The final resulting formula for the electrostatic potential due to a point charge +Q+Q at a distance rr is:     V=kQrV = \frac{kQ}{r}
  • Proportionality and Relationships:

    • Dependence on Charge (QQ):
    • The potential is directly proportional to the magnitude of the source charge:       VQV \propto Q
    • Dependence on Distance (rr):
    • Unlike the electric field which follows an inverse-square law (1r2\frac{1}{r^2}), the electrostatic potential follows an inverse relationship with distance:       V1rV \propto \frac{1}{r}
    • This implies that as the distance rr from the charge increases, the potential VV decreases linearly with the reciprocal of the distance.

Summary of Sidebar Identities and Rules

  • Integration Identities Used:
    • Rule 1: xndx=xn+1n+1\int x^{n} \, dx = \frac{x^{n+1}}{n+1}
    • Application: x2dx=x2+12+1=1x\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = -\frac{1}{x}
  • Mathematical Constants:
    • 1=0\frac{1}{\infty} = 0