Question

Using the symmetry of the arrangement, determine the direction of the force on q in the figure below, given that qa = qb = 8.55 ?C, qc = qd =

Answer 1

Given that q is a positive charge.Thus Qa and Qb will repell this q.Qc and Qd will attract this q as shown in the below figure.

The detailed figure is given below.

let the square have side length a.

a-8.55 q6 -8.55 2 2 lc -8.55 8.55

The force of repulsion between the charge q and Qa , q and Qb are same in magnitude and acts along the diagonal of the square as shown in the figure.

length of the diagonal of the square =$\sqrt{2}a$

let $q_{a}=q_{b}=Q$

hence we have $F_{a}=F_{b}=\frac{1}{4\pi \epsilon }\frac{Qq}{r^{2}}=\frac{1}{4\pi \epsilon }\frac{Qq}{(\frac{\sqrt{2}a}{2})^{2}}=\frac{1}{4\pi \epsilon }\frac{2Qq}{a^{2}}$

The resultant of this two forces is given by $F=\sqrt{F_{a}^{2}+F_{b}^{2}}=\sqrt{2}F_{a}$ since $F_{a}=F_{b}$

This resultant force will acts downwords.

Similarly Qc and Qd will attract q with a force of magnitude

$F_{c}=F_{d}=\frac{1}{4\pi \epsilon }\frac{Q'q}{r^{2}}=\frac{1}{4\pi \epsilon }\frac{Q'q}{(\frac{\sqrt{2}a}{2})^{2}}=\frac{1}{4\pi \epsilon }\frac{2Q'q}{a^{2}}$  ($q_{c}=q_{d}=Q'$)

the resultant attractive force equals $F=\sqrt{F_{c}^{2}+F_{d}^{2}}=\sqrt{2}F_{c}$ since $F_{c}=F_{d}$

this force will acts downwords.

Thus the net force acting on the charge q =$F=\sqrt{2}F_{a}+\sqrt{2}F_{c}=2\sqrt{2}F_{a}$

This force will be in the downward direction.

(since $q_{a}=q_{b}=-q_{c}=-q_{d}$ and the distance between these charges and q are the same we have $F_{a}=F_{b}=F_{c}=F_{d}$ )

b) Given $Q=8.55\mu C=8.55*10^{-6}C$ ,$q=2.35\mu C=2.35*10^{-6}C$,$a=10.90cm=10.90*10^{-2}m$

net force acting on q = $F=2\sqrt{2}F_{a}=2\sqrt{2}\frac{1}{4\pi \epsilon }\frac{2Qq}{a^{2}}=4\sqrt{2}*\frac{9*10^{9}*2.35*10^{-6}*8.55*10^{-6}}{(10.9*10^{-2})^{2}}$ =86.09N

downwards