Four point charges qA=2μC, qB=−5μC, qC=2μC, and qD=−5μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?

Question Id: PH0000007

Question: Four point charges qA = 2 µC, qB = –5 µC, qC = 2 µC, and qD = –5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?

Answer: zero 

Explanation: According to the given question you can draw the figure look like as given below.  

Fig-1: Coulomb's Force at the centre of square

Let, side of the square is a unit. Then length of the corner is \(a\sqrt{2}\) . So, the distance froom centre to corner is \(\frac{a}{\sqrt{2}}\) .

Given \(2\mu C\) charge will apply foce towards the centre. For both the \(2\mu C\) the applied forces on \(1\mu C\) chrge are equal and opposite to each other. And value of the force is

 \(F=\frac{1}{4 \pi \epsilon_0} \times \frac{1 \times 2}{(\frac{a}{\sqrt 2})^{2}} \mu N\)

\[\Rightarrow F=\frac{1}{4 \pi \epsilon_0} \times \frac{4}{a^{2}} \mu N\]

As the force are opposite and equal so  the resultant force due to both \(2\mu C\) charges is zero.

On the other hand applied force on \(1\mu C\) charge due to \(-5\mu C\) charge is outward from the centre. Again applied forces on \(1\mu C\) charge due to both  \(-5\mu C\) charge are equal and directed opposite to each other. And value of the force is

 \(F=\frac{1}{4 \pi \epsilon_0} \times \frac{1 \times (-5)}{(\frac{a}{\sqrt 2})^{2}} \mu N\)

\[\Rightarrow F=- \frac{1}{4 \pi \epsilon_0} \times \frac{10}{a^{2}} \mu N\]

As the force are opposite and equal so the resultant force due to both \(-5\mu C\) charges is zero.

As both resultant forces are zero so the resultant force of the whole system is zero.

Theory related to this problem:

• Condition for applying Coulomb's Law:

(i) Only applicable for point charges that means size of the point charge is negligible with respect to their distance.

(ii) Applicable for the distance larger than nuclear range ( \(\approx 10^{-15}m\) )

• Coulomb's law (Theory): Attraction force or repulsion force between between two point charge is proportional to the product of the charge and inversely proportional to the square of the distance between the charge.

i.e \(F\alpha (q_{1}\times q_{2})\)  and \(F\alpha \frac{1}{r^{2}}\)

So, \(F\alpha \frac{(q_{1}\times q_{2})}{r^{2}}\)

\(\Rightarrow F=k' \frac{(q_{1}\times q_{2})}{r^{2}}\)  ; [where k' is proportionality constant]

In C.G.S Unit: \(k'=\frac{1}{k}\)  [where k is electric permittivity]

Value of k=1 for air ( vaccum ) medium.

So, for C.G.S medium \(F=\frac{(q_{1}\times q_{2})}{r^{2}}\)

In S.I Unit: \(k'=\frac{1}{4\pi \epsilon }\)  [ \(\epsilon\) permittivity of medium]

And \(\epsilon =k\epsilon _{0}\)  where \(\epsilon _{0}\) is permittivity of air medium and k is dielectric constant.

So, \(F=\frac{1}{4\pi \epsilon _{0}}\times \frac{(q_{1}\times q_{2})}{r^{2}}\)

Value of  \(\epsilon _{0}=8.854\times 10^{-12} C^{2}N^{-1}m^{-2}\)

And value of  \(\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9} Nm^{2}C^{-2}\) 

Hence in S.I \(F=9\times 10^{9} \times \frac{(q_{1}\times q_{2})}{r^{2}}\) in N unit.

From vector form Coulomb's Law it can say that it is Newtonian in nature i.e \(\overrightarrow F_{12}=-\overrightarrow F_{21}\).