Question

Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium?

Answer 1

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

$q_1 = -Q_0$

$q_2 = -4Q_0$

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be $q_3 = +Q$ is placed at a distance x from $q_1$

$F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}$

The force on $q_3$ due to $q_2$ is

$F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}$

The condition of equilibrium is

$F_{1,3} = F_{2,3}$

$\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}$

$\frac{1}{x^2} = \frac{4}{(1-x)^2}$

$x = 0.331$ from $q_1$

To find the magnitude of $q_3$ we use $F_{1,2} = F_{1,3}$

$\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}$

$Q = 0.43Q_0$

The magnitude of the third charge must be 0.43 the first charge $Q_0$

Answer 2

The magnitude of the charge can be calculated by Coulomb's law.  The distance and magnitude of the third charge are 0.331 and 0.43  to the first charge.

To solve the given problem, use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's law.

The values for the two charges are:

$q_1 = -Q_0$

$q_2 = -4 Q$

   

As given in the question, both are negative charges, they will repulse each other. An equilibrium condition can be achieved only by placing the third Charge between the two

Assume, the third charge be $q_3 = +Q$   is placed at a distance $x$ from $q_1$  

$F_1,_3= \dfrac {k(-4Q_0)(+Q)}{x^2}$  

The force on$q_3$  due to $q_2$ is

$F_2,_3= \dfrac {k(-4Q_0)(+Q)}{1-x^2}$  

In the equilibrium condition,

$F_1_3 = F_2_3$

So,

$\begin {aligned} \dfrac {k(-4Q_0)(+Q)}{x^2} = \dfrac {k(-4Q_0)(+Q)}{1-x^2}\\x = 0.331 \rm \ From\ \it q_1 \end {aligned}$

       

The magnitude of $q_3$, can be calculated by

$F_1_2 = F_1_3$

So,

$\begin {aligned} \dfrac {k(-Q_0)(4Q)}{1^2} &= \dfrac {k(-Q_0)(Q)}{0.331^2}\\\\ Q &= 0.43 q_0\end {aligned}$

   

Therefore, the distance and magnitude of the third charge is 0.331 and 0.43  to the first charge.

To know more about the magnitude of charge,

https://brainly.com/question/13830003