Field of three charges in a row

Charges in a row on an axis. Three point charges on an axis, from left to right:

• $q_1 = +2\ \mu\text{C}$ at $x = 0$;
• $q_2 = -3\ \mu\text{C}$ at $x = 0{,}5\ \text{m}$;
• $q_3 = +5\ \mu\text{C}$ at $x = 1{,}0\ \text{m}$.

Calculate the electric field at points $A$ at $x = 0{,}25\ \text{m}$ and $B$ at $x = 0{,}75\ \text{m}$. Which has the larger magnitude?

Solution

Setting up. The total field is the vector sum (superposition principle) of the three contributions $E_i = \dfrac{k\,\abs{q_i}}{r_i^2}$. On an axis it is enough to manage the signs: take the $+x$ direction as positive. The field of a positive charge points away from it, that of a negative charge points towards it. Using $k = 9\cdot 10^9$.

Point A ($x = 0{,}25$ m). Distances: $r_1 = 0{,}25$, $r_2 = 0{,}25$, $r_3 = 0{,}75$. $\begin{aligned} E_1 &= \frac{(9\cdot 10^9)(2\cdot 10^{-6})}{0{,}25^2} = +2{,}88\cdot 10^5 \ (\text{from } q_1^+,\ \text{towards } +x) \\ E_2 &= \frac{(9\cdot 10^9)(3\cdot 10^{-6})}{0{,}25^2} = +4{,}32\cdot 10^5 \ (\text{towards } q_2^-\ \text{on the right},\ +x) \\ E_3 &= \frac{(9\cdot 10^9)(5\cdot 10^{-6})}{0{,}75^2} = -0{,}80\cdot 10^5 \ (\text{from } q_3^+,\ \text{towards } -x) \end{aligned}$ $E_A = (2{,}88 + 4{,}32 - 0{,}80)\cdot 10^5 \implies \ev{E_A \approx 6{,}4\cdot 10^5 \ \text{N/C} \ (\text{towards } +x)}$

Point B ($x = 0{,}75$ m). Distances: $r_1 = 0{,}75$, $r_2 = 0{,}25$, $r_3 = 0{,}25$. $\begin{aligned} E_1 &= \frac{(9\cdot 10^9)(2\cdot 10^{-6})}{0{,}75^2} = +0{,}32\cdot 10^5 \ (\text{from } q_1^+,\ \text{towards } +x) \\ E_2 &= \frac{(9\cdot 10^9)(3\cdot 10^{-6})}{0{,}25^2} = -4{,}32\cdot 10^5 \ (\text{towards } q_2^-\ \text{on the left},\ -x) \\ E_3 &= \frac{(9\cdot 10^9)(5\cdot 10^{-6})}{0{,}25^2} = -7{,}20\cdot 10^5 \ (\text{from } q_3^+,\ \text{towards } -x) \end{aligned}$ At B both $q_2$ (attracting to the left) and $q_3$ (repelling to the left) point in the same direction $-x$, so they add up: $E_B = (0{,}32 - 4{,}32 - 7{,}20)\cdot 10^5 \implies \ev{E_B \approx -1{,}1\cdot 10^6 \ \text{N/C} \ (\text{towards } -x)}$

Comparison. $\abs{E_B} \approx 1{,}1\cdot 10^6 > \abs{E_A} \approx 6{,}4\cdot 10^5$: the field is larger at B, because there the contributions of the negative charge $q_2$ and the large positive charge $q_3$ (both close, at $0{,}25$ m) agree in direction, whereas at A they partly cancel.

Connections

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