Ranking the field for various configurations

Ranking: field for various configurations. Four point charges at different distances:

• (A) $q = 1\ \mu\text{C}$ at $r = 0{,}1\ \text{m}$;
• (B) $q = 4\ \mu\text{C}$ at $r = 0{,}2\ \text{m}$;
• (C) $q = 9\ \mu\text{C}$ at $r = 0{,}3\ \text{m}$;
• (D) $q = 16\ \mu\text{C}$ at $r = 0{,}4\ \text{m}$.

Calculate and rank the field $E = kq/r^2$ at the respective points. What do you notice about the four values, and why?

Solution

Setting up. We apply $E = \dfrac{kq}{r^2}$ directly to each configuration, with $k = 9\cdot 10^9$. $\begin{aligned} E_A &= \frac{(9\cdot 10^9)(1\cdot 10^{-6})}{(0{,}1)^2} = \frac{9\cdot 10^3}{0{,}01} = 9\cdot 10^5 \ \text{N/C} \\ E_B &= \frac{(9\cdot 10^9)(4\cdot 10^{-6})}{(0{,}2)^2} = \frac{3{,}6\cdot 10^4}{0{,}04} = 9\cdot 10^5 \ \text{N/C} \\ E_C &= \frac{(9\cdot 10^9)(9\cdot 10^{-6})}{(0{,}3)^2} = \frac{8{,}1\cdot 10^4}{0{,}09} = 9\cdot 10^5 \ \text{N/C} \\ E_D &= \frac{(9\cdot 10^9)(16\cdot 10^{-6})}{(0{,}4)^2} = \frac{1{,}44\cdot 10^5}{0{,}16} = 9\cdot 10^5 \ \text{N/C} \end{aligned}$ Result. $\ev{E_A = E_B = E_C = E_D = 9\cdot 10^5 \ \text{N/C}}$ Why. This is no coincidence: in each configuration the charge grows as the square of the distance, $q \propto r^2$ (the values $1,4,9,16$ are $r$ measured in decimetres, squared). In the ratio $E = kq/r^2$ the factor $r^2$ cancels and the field stays identical. It is the complement of the inverse-square law: doubling the distance drops the field by 4, but quadrupling the charge brings it back to the initial value.

Connections

Topics: Electric field and potential Concepts: Electric field · Coulomb’s law Skills: Superposition principle Methods: Gradient method Objects: Charged sphere