91影视

The values of charges forming a square are: $q_1(=q_4)=Q$and $q_2(=q_3)=q$

Using the dot product of the charges, we can get the value of the force. Using this value and the condition of the net force on particles 1 and 4 to be zero, we can find the required ratio of the charges. Again, using the concept for the y-component of particle 2, we can observe the equilibrium structure of the conditions.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=\frac{1}{\left(4蟺{蔚}_0\right)}\left(\frac{\left|q_1\right|\left|q_2\right|cos胃}{r^2}\stackrel{铃�}{i}+\frac{\left|q_1\right|\left|q_2\right|\sin 胃}{r^2}\stackrel{铃�}{j}\right)$ (1)

For ease of presentation (of the computations below), we assume Q > 0 and

q< 0 (although the final result does not depend on this particular choice).

The x-component of the force experienced by q₁ =Q using equation (1) is given as:

$$\begin{gathered} {F_1}_x=\frac{1}{4蟺{蔚}_0}\left(\frac{-\left(Q\right)\left(Q\right)}{(鈭�2a{)}^2}cos45掳+\frac{\left(\right|q\left|\right)\left(Q\right)}{a^2}\right) \\ {F_1}_x=\frac{Q\left|q\right|}{4蟺{蔚}_0{a}^2}\left(\frac{-(Q/|q|}{(2鈭�2)}+1\right) \\ \frac{Q\left|q\right|}{4蟺{蔚}_0{a}^2}\left(-\frac{Q/\left|q\right|}{2鈭�2}+1\right)=0 \end{gathered}$$

Therefore,

role="math" localid="1662710310471" $$\begin{gathered} Q/\left|q\right|=2\sqrt{2} \\ =2.83 \end{gathered}$$

Or, it can be,

$$\begin{gathered} Q/\left|q\right|=-2\sqrt{2} \\ =-2.83 \end{gathered}$$

Thus, the value of the ratio is 2.83 and -2.83.

The y-component of the net force on q₁=Q using equation (1) is given as:

$\begin{array}{l}{F}_{2y}=\frac{1}{4{\mathrm{蟺蔚}}_0}\left(\frac{{\left|q\right|}^2}{({\sqrt{2a)}}^2}\sin {45}^{鈭�}-\frac{\left|q\right|\left(Q\right)}{a^2}\right)\\ F_{2y}=\frac{|q{|}^2}{4{\mathrm{蟺蔚}}_0{a}^2}\left(\frac{1}{2\sqrt{2}}-\frac{\left(Q\right)}{\left|q\right|}\right)\end{array}$

$(ifweconsidertheconditionofnetzeroforceoneachcharge,then)Q/q=(-1)/(2鈭�2)$

The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1.

Hence, there is no value of q.