Answer to Question #177924 in Physics for Bernadette

Question

4. At the corners of a rectangle ABCD of sides 8 cm and 6 cm are charges of +50μC, -30μC, +20μC 

and -40μC, respectively. AB is longer side and BC is the shorter side. Find the force (in N) 

on a +15μC charge placed at the center of the rectangle

Answer 1

Let $q_1 = +50\mu C, q_2 = -30\mu C, q_3 = +20\mu C, q_4 = -40\mu C, q = +15\mu C$, and $AB = 8cm = 0.08m, BC = 6cm = 0.06m$.

Then, according to the Pyphagorean theorem:

$$AO = BO = CO = DO = \sqrt{\left( \dfrac{AB}{2} \right)^2 + \left( \dfrac{AC}{2} \right)^2} = \sqrt{0.04^2 + 0.03^2} = 0.05m$$

Then, according to the Coulomb's law, the force on $q$ from $q_1$ is:

$$F_1 =k \dfrac{|q||q_1|}{AO^2}$$

where $k\approx 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant.

Thus, find:

$$F_1 = 9\times 10^9\cdot \dfrac{15\times 10^{-6}\cdot 50\times 10^{-6}}{0.05^2} = 2.7\times 10^{3}N$$

The force on $q$ from $q_2$ is:

$$F_2 =k \dfrac{|q||q_2|}{BO^2} = 1.62\times 10^3N$$

The force on $q$ from $q_3$ is:

$$F_3 =k \dfrac{|q||q_3|}{BO^2} = 1.08\times 10^3N$$

The force on $q$ from $q_4$ is:

$$F_4 =k \dfrac{|q||q_3|}{BO^2} = 2.16\times 10^3N$$

The directions of these forces are show in figure above (like charges repel each other; unlike charges attract).

Adding the forces along $AD$ and $BC$ obtain:

$$F_{AD} = F_1-F_3 = 1.62\times 10^{3}N$$ $$F_{BC} = F_4-F_2 = 0.54\times 10^{3}N$$

These forces and their sum $F$ are shown in figure below

According to the cosine law, the length of vector $F$ is:

$$F = \sqrt{F_{BC}^2 + F_{AD}^2 - 2F_{BC}F_{AD}\cos\angle BOD}$$

The angle $\angle BOD$ is:

$$\angle BOD = 2\arccos\left( \dfrac{OE}{OB} \right) = 2\arccos\left( \dfrac{0.04}{0.05} \right) = 2\arccos0.8$$

Substituting these values, obtain:

$$F = \sqrt{(0.54\times 10^3)^2 + (1.62\times 10^3)^2 - 2\cdot 0.54\cdot 1.62\times 10^6\cdot 0.28}\approx 1.56\times 10^3N$$

Answer. $1.56\times 10^3N$.