Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find  the
(a) resultant electric force on a charge Q, and
(b) potential energy of this system.

 

 

 

 
 
 
 
 

Answers (1)
S safeer

a)  
      
Force on charge Q at C is the resultant force of all other charges.
By Coulomb's law force between two charges is,
F= \frac{1}{4\pi \varepsilon _{0}}\: \frac{q_{1}q_{2}}{r^{2}}
where q1 and q2 are charges separated by distance 'r'
By applying this law, the force due to charge q is
By q at D,
               F_{1}= \frac{1}{4\pi \varepsilon _{0}}\: \frac{qQ}{a^{2}}
By q at B, F_{2}= \frac{1}{4\pi \varepsilon _{0}}\: \frac{qQ}{a^{2}}
where 'a' is the distance between charges
The distance between charge Q at C and Q atA is ,\sqrt{2}\: a
So, by, Applying Coulomb's law, we get
      F_{3}= \frac{1}{4\pi \varepsilon _{0}}\; \frac{Q^{2}}{{2}\, a^{2}}

The resultant of force F1 and F2 is along F3
That is , F_{12}=\sqrt{F_{1}^{2}+F_{2}^{2}}= \sqrt{2}\, F
                    =\frac{\sqrt{2}}{4\pi \varepsilon _{0}}\; \frac{qQ}{a^{2}}
Now, the net force is F_{net}=F_{3}+F_{12}
                                        = \frac{\sqrt{2}}{4\pi \varepsilon _{0}}\: \frac{qQ}{a^{2}}+\frac{1}{4\pi \varepsilon _{0}}\, \frac{Q^{2}}{\left ( \sqrt{2} \, a\right )^{2}}
                                       = \frac{Q}{4\pi \varepsilon _{0}}\left [ \frac{\sqrt{2}q}{a^{2}}+\frac{Q}{2a^{2}} \right ]
                                      = \frac{Q}{4\pi \varepsilon _{0}}\left [ \frac{2\sqrt{2}q+Q}{2a^{2}} \right ]
The direction of net force is along AC
b) The potential energy between two charges separated by a distance r is
\cup = \frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r}
For finding the potential energy of a system, first, find separate potential energy.
The potential energy between charges which are near corners is (separated by a)
\cup _{1} =4\times \frac{1}{4\pi \varepsilon _{0}}\, \cdot \frac{qQ}{a}
The potential energy between charge q and q is
\cup _{2} = \frac{1}{4\pi \varepsilon _{0}}\, \cdot \frac{q^{2}}{\sqrt{2}\, a}
The potential energy between charge Q and Q is
\cup _{3} = \frac{1}{4\pi \varepsilon _{0}}\, \cdot \frac{Q^{2}}{\sqrt{2}\, a}
Therefore, the net potential energy of a system is
\cup = \cup _{1}+ \cup _{2}+ \cup _{3}
   = 4\times \frac{1}{4\pi \varepsilon _{0}}\: \cdot \tfrac{qQ}{a}+\frac{1}{4\pi \varepsilon _{0}}\, \frac{q^{2}}{\sqrt{2}\, a}+\frac{1}{4\pi \varepsilon _{0}}\, \frac{Q^{2}}{\sqrt{2}a}
= \frac{1}{4\pi \varepsilon _{0}a}\left [ 4qQ+\frac{q^{2}}{\sqrt{2}}+\frac{Q^{2}}{\sqrt{2}} \right ]

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