Question

Four-point charges q1 = 2 micro C, q2 = -5 micro C, q3 = 2 micro C, and q4 = -5 micro Care located add the corners of a square ABCD of side 10 cm. what is the force on a charge of1 micro C placed at the centre of square?​

Answer 1

$\underline{\blacksquare\:\:\:\footnotesize{\purple{\textsf{ \textbf{{SOLUTION:-}}}}}}$

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• Side of square (a) = 10 cm = 0.1 m

• Five point charges :

1. q₁ = 2μC = 2 × 10⁻⁶ C
2. q₂ = -5μC = 5 × 10⁻⁶ C
3. q₃ = 2μC = 2 × 10⁻⁶ C
4. q₄ = -5μC = 5 × 10⁻⁶ C
5. q₅ = 1 μC = 1 × 10⁻⁶ C (Point charge present at the centre)

• Half of the diagonal of the square = a/√2 = 0.1/√2 m

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See the attached figure!

• First of all we will find the electric field along diagonal AC due to point charge q₁ = 2μC and q₃ = 2μC.

• Due to point Charge q₁ = 2μC, the electric field will be away from the centre.

• Due to point Charge q₃ = 2μC, the electric field will be away from the centre.

$\footnotesize{\longrightarrow \rm E_{(AC)}=E_{1} - E_{3} }$

$\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{Kq_{1} }{r^2} - \dfrac{Kq_{3} }{r^2} }$

$\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{9 \times {10}^{9} \times 2 \times {10}^{ - 6} }{ \bigg( \frac{0.1}{ \sqrt{2} } \bigg)^{2} } - \dfrac{9 \times {10}^{9} \times 2\times {10}^{ - 6} }{ \bigg(\frac{0.1}{ \sqrt{2} } \bigg)^{2}} }$

• The magnitude of the charges are equal therefore they will cancel each other and electric field along diagonal AC will be 0.

$\footnotesize{\longrightarrow \underline{ \boxed{ \red{\bf E_{(AC)} = 0 \: N/C}}}}$

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•Now, let's find the electric field along diagonal BD.

• Due to point charge q₂ = -5μC ,the electric field will be directed towards the centre.

• Due to point charge q₄ = -5μC, the electric field will be directed towards centre.

• From the observation we find that the magnitude of the charges are equal therefore they will cancel each other.

• Hence, electric field along diagonal BD will be 0.

$\footnotesize{\longrightarrow \underline{ \boxed{ \orange{\bf E_{(BD)} = 0 \: N/C}}}}$

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At the centre there are two electric field which are perpendicular to each other, so net electric field can be calculated by using superposition principle:

$\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} + 2 E_{(AC)} ^{2} E_{(BD)}^{2} \cos( {90}^{ \circ} ) }}$

$\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} }}$

$\footnotesize{\longrightarrow \rm E_{(Net)} =E \sqrt{2 }}$

$\footnotesize{\longrightarrow \underline{ \boxed{ \gray{\bf E_{(Net)} =0 \: N/C}}}}$

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