Question

Four point charges qA = -2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are located at the corners of a square ABCD with 10 cm sides. What is the force on a charge of 1 μC placed at the centre of the square? pls ans fast urgent​

Answer 1

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

i) The four point charges

• -2μC or -2 × (10)^6 C
• - 5 μC or -5 × (10)^6 C
• 2 μC or 2 × (10)^6 C
• -5 μC or -5 × (10)^6 C are located at the corners of square ABCD

ii) The sides of the square is 10 cm or 0.1 m , So , half of the diagonal is 0.1/√2 m

We know that , The electrostatic force between two point charges is given by

$\huge{\mathtt{ \fbox{F = k\frac{ q_{1}q_{2}}{ {(r)}^{2} }} }}$

Thus ,

The force on O due to A is

$\sf \mapsto F_{1} = k \frac{(2)\times 1}{ {( \frac{0.1}{ \sqrt{2} }) }^{2} } \\ \\ \sf \mapsto F_{1}= \frac{4k}{0.01} \: \: newton$

The force on O due to C is

$\sf \mapsto F_{2} = k \frac{(2)\times 1}{ {( \frac{0.1}{ \sqrt{2} }) }^{2} } \\ \\ \sf \mapsto F_{2} = \frac{4k}{0.01} \: \: newton$

and the force on O due B and D are equal in magnitude but opposite in directions thus , they will cancel each other

Hence , the resultant force of F1 and F2 is

$\star \: \: \: \large \mathtt{ \fbox{Resultant \: force =F_{1} + F_{2}}} \\ \\ \sf \mapsto Resultant \: force = \frac{4k}{0.01} + \frac{4k}{0.01} \\ \\ \sf \mapsto Resultant \: force = \frac{8k}{0.01} \\ \\\sf \mapsto Resultant \: force = \frac{8 \times 9 \times {(10)}^{9} }{0.01} \\ \\\sf \mapsto Resultant \: force = 72 \times {(10)}^{(9 + 2)} \\ \\\sf \mapsto Resultant \: force = 72 \times {(10)}^{11} \: \: newton$

The electrostatic force on the centre of square is 72 × (10)^11 newton

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