Question

Four charges on the corner of square q1=q2=+1µc on the top of the square andq3=q4=-2µc on the bottom of the square, Find the resultant force on q4

Answer 1

Explanations & Calculations

• Refer to the figure attached

• F1 is repulsive due to the like q3 & q4 charges & both F2, F3 are attractive due to the opposite q1, q2 charges.
• Once those 3 forces are found (by Coulomb's law), the resultant generated on q4 can be calculated by resolving them into orthogonal 2 components.
• That is the idea behind this question.
• You need to know the length of a side of the square. If it is $\small a$ for general case, then the diagonal length of the square is $\small \sqrt2a$
• Then,

$\qquad\qquad \begin{aligned} \small F_1&=\small k\frac{q_3q_4}{a^2}=k\frac{(2\times10^{-6}C\times 2\times 10^{-6}C)}{a^2}\\ &=\small 4\bigg[\frac{10^{-12}k}{a^2}\bigg]\,N\\\\ \small F_2&=\small k\frac{(2\times10^{-6}C\times1\times10^{-6}C)}{a^2}\\ \small&=\small 2\bigg[\frac{10^{-12}k}{a^2}\bigg]\,N\\\\ \small F_3&=\small k\frac{(2\times10^{-6}C\times1\times10^{-6}C)}{2a^2}\\ &=\small \bigg[\frac{10^{-12}k}{a^2}\bigg]\,N \end{aligned}$

• Then the orthogonal components are,

$\qquad\qquad \begin{aligned} \small Y&=\small F_2+F_3\sin45\\ &=\small \bigg[\frac{10^{-12}k}{a^2}\bigg]\bigg(2+\frac{1}{\sqrt2}\bigg)\\ &=\small 2.707\bigg[\frac{10^{-12}k}{a^2}\bigg]\,N\\\\ \small X&=\small F_1-F_3\cos45\\ &=\small 3.293\bigg[\frac{10^{-12}k}{a^2}\bigg]\,N \end{aligned}$

• Then the resultant is,

$\qquad\qquad \begin{aligned} \small R&=\small \sqrt{X^2+Y^2}\\ &=\small \bigg[\frac{10^{-12}k}{a^2}\bigg]\sqrt{(2.707)^2+(3.293)^2}\\ &=\small 4.263\bigg[\frac{10^{-12}k}{a^2}\bigg]\,N \end{aligned}$

• Substituting the values of $\small a$ (should be in meters) and $\small k$ (= $\small 9\times10^{-9}Nm^2C^{-2}$ ), the resultant force could be found.