Question

Three charges of each Q are placed at the vertices of a square of side 1m calculate the force on -q charge at the centre placed of square

Answer 1

Answer:

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Answer 2

Explanation:

Force due to charge on corner A=

$\frac{1}{4\piϵ} \frac{ {q}^{2} }{ \frac{a}{2} {}^{2} } oa$

Force due to charge on corner B=

$\frac{1}{4\piϵ} \frac{ {q}^{2} }{ \frac{a}{2} {}^{2} } ob \:$

Force due to charge on corner C=

$\frac{1}{4\piϵ} \frac{ {q}^{2} }{ \frac{a}{2} {}^{2} }$

Total force=

$\frac{1}{4\piϵ} \frac{ {q}^{2} }{ \frac{a}{2} {}^{2} } = \frac{1}{2\piϵ} \frac{ {q}^{2} }{ {a}^{2} } oc$