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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.
Solution
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,
\[F' = 14 . 4\sqrt{2} + 7 . 2\]
\[ = 27 . 56 N\]
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