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Three equal charges, 2⋅0 × 10 − 6 C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.
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Asked by ratnadeep.dmr003 | 21 Apr, 2024, 11:06: PM
answered-by-expert Expert Answer

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Let OAB is the equilateral triangle of side 5 cm . Let each vertex of triangle has charge 2 μC .

Vertex O is at origin and side OA be on x-axis.

Force FA is electrostatic force on charge at O due to charge at A . Force FB is electrostatic force on charge at O due to charge at B .

Electrostatic force F between charges q1 and q2 that are separated by a distance d is

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where K = 9 × 109 N m2 C-2 is Coulomb's constant

Magnitudes of forces FA and FB are equal because distance between charges are equal.

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we get, |FA | = | FB | = 14.4 N

Net force on charge at O is

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Magnitude of force is

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Hence each charge at the vertex of triangle experiences a force 14.4 N due to other charges

 

 

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