Today’s question 💭: Imagine a single point charge. How does it fill the space around it? Is the field stronger near it or farther away? And which direction does the field point? Let’s derive the answer directly from Coulomb’s law 👇

Building the Formula Ourselves 🛠️

We want to find the electric field produced by a point charge $q$ at a distance $r$ from it. It’s simple: we place a test charge $q_0$ there. According to Coulomb’s law, the force acting on it is:

$$ F = kfrac{|q||q_۰|}{r^۲} $$

Now, we use the definition of the electric field ($E = F/q_0$). We just need to divide both sides by $q_0$; that $q_0$ cancels out, leaving us with:

$$ E = kfrac{|q|}{r^۲} tag{۱-۴} $$

This formula tells us two things: the field is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance. In other words, if you double the distance, the field becomes one-fourth as strong. ✨

Direction of the Field: Which Way? 🧭

Remember that the direction of the field is the direction of the force exerted on a positive test charge. So:








+

Positive charge field: pointing outward




−
Explore This Field Yourself 🎮

In this simulation, make the charge positive or negative, adjust its magnitude, and drag the blue point (sensor) to see how the field strength varies at different distances. Notice how the arrows get shorter (indicating a weaker field) as you move farther away:

Van de Graaff Generator Example 🔮

💡 Example 1-6 (from the textbook): The dome of a Van de Graaff generator has accumulated a charge of $1.0 mutext{C}$. What is the electric field at distances of 1, 2, 3, and 4 meters from the center? (Using $E = k|q|/r^2$):
– $E_1 = dfrac{(9times10^9)(1.0times10^{-6})}{(1)^2} = 9.0times10^3 text{N/C}$
– $E_2 = dfrac{(9times10^9)(1.0times10^{-6})}{(2)^2} approx 2.3times10^3 text{N/C}$
– $E_3 approx 1.0times10^3 text{N/C}$ , $E_4 approx 0.56times10^3 text{N/C}$
Observe how the field drops to approximately one-fourth when the distance doubles (1→2). This is the inverse square law.

A neat experiment: place two candles near a Van de Graaff generator — one close, one far. The flame of the near candle is clearly pulled toward the dome (because the negative dome attracts the positive ions in the flame), but the flame of the far candle barely moves — because it is in a weaker field. 🕯️

When We Have Multiple Charges: Superposition of Fields ➕

Like forces, fields are also additive. The net field at a point is the vector sum of the fields from each individual charge (as if the others weren’t there):

$$ vec{E} = vec{E}_۱ + vec{E}_۲ + cdots tag{۱-۵} $$

This is the principle of superposition of electric fields. In this book, examples are always set up so that the fields are either along the same line or perpendicular to each other — so the vector sum stays simple. 😌

🧮 Calculate it yourself (Exercise 1-4): According to the Bohr model, the distance of the electron from the proton in the hydrogen nucleus is $5.3times10^{-11} text{m}$. a) What is the field due to the proton at this distance? b) At what distance from the proton does the field equal the field of the Van de Graaff generator from the example above at a distance of 1 meter (i.e., $9.0times10^3 text{N/C}$)?

Informal Summary 🎁

  • Point charge field: $E = kdfrac{|q|}{r^2}$ — directly proportional to charge, inversely proportional to the square of the distance.
  • Direction: positive points outward, negative points inward.
  • Multiple charges? Net field = vector sum (superposition).

🔗 For the Curious — Deeper Study


Test Yourself 📝

Click on each question to reveal the answer 👇


In the next section (1-6), we’ll explore a fantastic visual way to see the field: electric field lines. 👋

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