Question

Four point charges -q, -q, +q & +q are located at the vertices A, B, C & D of a square of side 'a' respectively. Find the net force on +q' located at C due to

rest of three charges.


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Answer 1

☆Coulomb Force

》It measure the electrostatic forces experienced by two charge particles kept at some distance 'r'.

》It is a vector quantity, magnitude gives the amount of force; while direction gives the nature of forces(attractive/ repulsive)

$\boxed{\bf{ \vec{F} = \dfrac{ kq_1 q_2}{r²}}}$

☆Solution

see the above figure.

• Arrows of forces show the repulsion(like charges) and attraction(unlike charge) nature.

• $\rm \vec{F}_{C} = \vec{F}_{A} +\vec{F}_{B}+\vec{F}_{D}$

• $\rm \vec{F}_{C} = \dfrac{kq²}{a²}\hat{j}+ \dfrac{kq²}{(\sqrt{2}a)²}sin45°\hat{j}+\dfrac{kq²}{(\sqrt{2}a)²}cos45°\hat{i}+ \dfrac{kq²}{a²}\hat{-i}$

•$\rm \vec{F}_{C} = (\dfrac{kq²}{a²}+ \dfrac{kq²}{2\sqrt{2}a²})\hat{j}+(\dfrac{kq²}{2\sqrt{2}a²}– \dfrac{kq²}{a²})\hat{i}$

• $\bf \vec{F}_{C} = \dfrac{kq²}{a²}(1+\dfrac{1}{\sqrt{2}})\hat{i} + \dfrac{kq²}{a²}(1-\dfrac{1}{\sqrt{2}})\hat{j}$

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