Understanding Electric Field Magnitude from a Point Charge

This question asks for the formula describing the magnitude of the electric field at a certain distance r from a point charge Q.

Definition of Electric Field

An electric field represents the force exerted per unit positive charge in a region of space around an electric charge or distribution of charges. It's a vector quantity, indicating both strength and direction. The magnitude quantifies how strong the field is at a particular point.

Basis in Coulomb's Law

The calculation of the electric field due to a point charge relies on Coulomb's Law, which defines the electrostatic force ($F$) between two point charges ($q_1$ and $q_2$) separated by a distance ($r$):

$$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}$$

In this formula:

• $F$ is the magnitude of the electrostatic force.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the separation distance between the charges.
• $\varepsilon_0$ is the permittivity of free space ($ \approx 8.854 \times 10^{-12} \, \text{F/m}$).
• The term $\frac{1}{4\pi\varepsilon_0}$ is Coulomb's constant, often denoted by $k$.

Deriving the Electric Field Magnitude Formula

The electric field ($E$) at a point is defined as the force ($F$) experienced by a tiny positive test charge ($q_0$) placed at that point, divided by the magnitude of the test charge itself:

$$E = \frac{F}{q_0}$$

Consider a source point charge Q. To find the electric field magnitude at a distance r from Q, we place a test charge $q_0$ at that distance. The force exerted by Q on $q_0$ is given by Coulomb's Law:

$$F = \frac{1}{4\pi\varepsilon_0} \frac{|Q q_0|}{r^2}$$

Now, substitute this force into the definition of the electric field:

$$E = \frac{F}{q_0} = \frac{1}{q_0} \left( \frac{1}{4\pi\varepsilon_0} \frac{|Q q_0|}{r^2} \right)$$

The test charge $q_0$ cancels out, leaving the formula for the electric field magnitude due to the source charge Q:

$$E = \frac{1}{4\pi\varepsilon_0} \frac{|Q|}{r^2}$$

The magnitude of the electric field depends directly on the magnitude of the source charge ($|Q|$) and inversely on the square of the distance ($r^2$) from the charge.

Evaluating the Options

Let's compare the derived formula $E = \frac{1}{4\pi\varepsilon_0} \frac{|Q|}{r^2}$ with the given options:

• Option 1: $E = \frac{Q}{4\pi\varepsilon_0 r}$ presents an incorrect distance dependence ($1/r$ instead of $1/r^2$).
• Option 2: $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ accurately reflects the derived formula, showing the correct dependence on $Q$ and $r^2$.
• Option 3: $E = \frac{Q^2}{4\pi\varepsilon_0 r^2}$ incorrectly states the field magnitude depends on the square of the charge ($Q^2$) instead of being directly proportional to $Q$.
• Option 4: $E = \frac{Q}{2\pi\varepsilon_0 r^2}$ has the correct distance dependence ($1/r^2$) but uses an incorrect constant factor ($1/(2\pi\varepsilon_0)$ instead of $1/(4\pi\varepsilon_0)$).

Final Answer Confirmation

Based on the derivation from Coulomb's Law, the correct expression for the magnitude of the electric field at a distance r from a point charge Q is $E = \frac{Q}{4\pi\varepsilon_0 r^2}$.