Question

deux charges ponctuelles, a = -5 q et b = 3 q, sont situées aux positions montrées dans la figure ci - dessus.
a) quelle est la force sur une charge +q à lorigine?
□i+□j kqq n
b) où placeriez - vous une charge ponctuelle +4.5 q de sorte que la force nette sur q soit nulle?
(x,y)=(□,□)

Explanation:

Step1: Calculate force due to A

The distance of charge A from origin $r_A = 3m$. Using Coulomb's law $F = k\frac{Q_1Q_2}{r^2}$, the force due to A on $+q$ is $F_A=k\frac{(- 5Q)q}{3^2}=-\frac{5kQq}{9}\hat{i}$.

Step2: Calculate force due to B

The distance of charge B from origin $r_B = 2m$. The force due to B on $+q$ is $F_B=k\frac{(3Q)q}{2^2}=\frac{3kQq}{4}\hat{j}$.

Step3: Find net force

The net force $F = F_A+F_B=-\frac{5kQq}{9}\hat{i}+\frac{3kQq}{4}\hat{j}$.

For part b:
Let the position of $+4.5Q$ be $(x,y)$. The net force on $q$ is zero when the force due to $+4.5Q$ balances the sum of forces due to A and B. Since the net - force in x - direction due to A and B is $-\frac{5kQq}{9}$ and in y - direction is $\frac{3kQq}{4}$.
Let's assume the charge $+4.5Q$ is placed such that its x - component of force cancels $F_{Ax}$ and y - component cancels $F_{By}$. By symmetry and Coulomb's law, we consider the x - and y - components separately.
The force due to $+4.5Q$ on $q$ is $F_{4.5Q}=k\frac{(4.5Q)q}{r^2}$. Let's assume it is placed on the line joining the resultant of $F_A$ and $F_B$.
We know that for the net force to be zero, considering the x - direction:
Let the distance of $+4.5Q$ from the origin in x - direction be $x$ and in y - direction be $y$.
The force due to $+4.5Q$ in x - direction $F_{4.5Qx}=k\frac{(4.5Q)q}{x^2 + y^2}\frac{x}{\sqrt{x^2 + y^2}}$ and in y - direction $F_{4.5Qy}=k\frac{(4.5Q)q}{x^2 + y^2}\frac{y}{\sqrt{x^2 + y^2}}$.
Equating the x - components: $k\frac{(4.5Q)q}{x^2 + y^2}\frac{x}{\sqrt{x^2 + y^2}}=\frac{5kQq}{9}$ and for y - components: $k\frac{(4.5Q)q}{x^2 + y^2}\frac{y}{\sqrt{x^2 + y^2}}=\frac{3kQq}{4}$.
Dividing the two equations $\frac{y}{x}=\frac{\frac{3}{4}}{\frac{5}{9}}=\frac{27}{20}$.
Let's assume the charge $+4.5Q$ is placed on the line passing through the origin and in the opposite direction of the net - force of A and B on $q$.
The net force due to A and B on $q$ has direction $\theta=\tan^{-1}(\frac{F_y}{F_x})=\tan^{-1}(-\frac{27}{20})$.
We can also use the fact that the magnitude of the force due to $+4.5Q$ should be equal to the magnitude of the net - force of A and B on $q$.
The magnitude of the net - force of A and B on $q$ is $F_{net}=\sqrt{(\frac{5kQq}{9})^2+(\frac{3kQq}{4})^2}$.
And $F_{4.5Q}=k\frac{(4.5Q)q}{r^2}=F_{net}$.
After solving (equating the magnitudes and using the ratio of components), we find that the charge $+4.5Q$ should be placed at a point such that the x - and y - components of the force due to it cancel the x - and y - components of the net - force of A and B on $q$.
The position $(x,y)$ should satisfy the condition that the force due to it balances the net - force of A and B on $q$.
By equating the forces and solving we get $(x,y)=( - 1.8, - 2.43)$.

Answer:

a) $-\frac{5kQq}{9}\hat{i}+\frac{3kQq}{4}\hat{j}$
b) $(-1.8,-2.43)$