Question

Four equal charges of 2.0 × 10⁻⁶ C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb's force experienced by one of the charges due to the other three. 

Answer 1

Given,
Magnitude of the charges, 

\[q = 2 \times  {10}^{- 6}   C\] Side of the square,

\[a = 5  \text{ cm = 0 . 05  m }\]

By Coulomb's Law, force, 

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]     

  

So, force on the charge at A due to the charge at B,

\[\vec{F}_{BA}  = \frac{9 \times {10}^9 \times \left( 2 \times {10}^{- 6} \right)^2}{\left( 0 . 05 \right)^2}      \]  

\[             = \frac{9 \times {10}^9 \times 4 \times {10}^{- 12}}{25 \times {10}^{- 4}}\] 

\[             = 14 . 4  N\] 

Force on the charge at A due to the charge at C,

\[\vec{F}_{CA}  = \frac{9 \times {10}^9 \times \left( 2 \times {10}^{- 6} \right)^2}{\left( \sqrt{2} \times 0 . 05 \right)^2}\] 

\[             = \frac{9 \times {10}^9 \times 4 \times {10}^{- 12}}{25 \times 2 \times {10}^{- 4}}\] 

\[             = 7 . 2  N\] 

Force on the charge at A due to the charge at D,

\[\vec{F}_{DA}  =  \vec{F}_{BA}\]

The resultant force at A, F'= \[\vec{F}_{BA}  +  \vec{F}_{CA}  +  \vec{F}_{DA}\]

The resultant force of \[\vec{F}_{DA}  \text{ and  } \vec{F}_{BA}\]  will be \[\sqrt{2} F_{BA}\]  in the direction of \[\vec{F}_{CA}\] . Hence, the resultant force,

\[F' = 14 . 4\sqrt{2} + 7 . 2\] 

\[       = 27 . 56  N\]