Question

1) find the resultant force at o​

Answer 1

Answer:

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• Side of square (a) = 3 cm = (3/100) = 0.03 m

• Half of the diagonal = (a/√2) = (0.03/√2)

• Magnitude of charge :

• q₁ = + 6 nC = 6 × 10⁻⁹ C
• q₂ = +8 nC = 8 × 10⁻⁹ C
• q₃ = -4 nC = 4 × 10⁻⁹ C
• q₄ = +3 nC = 3 × 10⁻⁹ C

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Electric field along diagonal BD :

•Due to charge q₂ it's electric field is directed towards point D i.e away from the centre.

• Due to charge q₄ it's electric field is directed towards point B i.e away from the centre. Therefore, electric field along diagonal BD is :

$\footnotesize{\longrightarrow\: \sf E_{(BD)}=E_{2} -E_{1} }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{KQ}{ {a}^{2} } - \dfrac{KQ}{ {a}^{2} } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{9 \times {10}^{9} \times8 \times {10}^{ - 9} }{\bigg( \frac{0.03}{\sqrt{2} } \bigg)^{2} } - \dfrac{9 \times {10}^{9} \times3 \times {10}^{ - 9} }{ \bigg( \frac{0.03}{\sqrt{2} } \bigg)^{2} } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{72}{\frac{0.0009}{2 } } - \dfrac{ 27 }{ \frac{0.0009}{2 } } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{72 \times 2}{0.0009 } - \dfrac{ 27 \times 2}{ 0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{144}{0.0009 } - \dfrac{54 }{ 0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{144 - 54}{0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{90}{0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{90 \times 10000}{0.0009 \times 10000 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= \dfrac{900000}{9 } }$

$\footnotesize{\longrightarrow\: \sf E_{(BD)}= 100000 \: N/C}$

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Electric field along AC :

• Due to charge q₁ it's electric field will be directed away from the centre i.e directed towards point C.

• Due to charge q₃ it's electric field will be towards centre and directed towards point C.

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=E_{2} + E_{1} }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=\dfrac{KQ}{ {a}^{2} } + \dfrac{KQ}{ {a}^{2} } }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=\dfrac{9 \times {10}^{9} \times6 \times {10}^ { - 9} }{ \bigg( \frac{0.03}{\sqrt{2} } \bigg)^{2} } + \dfrac{9 \times {10}^{9} \times 4 \times {10}^{ - 9} }{ \bigg( \frac{0.03}{\sqrt{2} } \bigg)^{2}} }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=\dfrac{108 }{ 0.0009 } + \dfrac{72 }{ 0.0009} }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=\dfrac{108 + 72 }{ 0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=\dfrac{180 }{ 0.0009 } }$

$\footnotesize{\longrightarrow\: \sf E_{(AC)}=200000 \: N/C}$

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

$\footnotesize{\longrightarrow\: \sf E_{(net)}=\sqrt{(E_{AC})^2+ (E_{BD})^2}}$

$\footnotesize{\longrightarrow\: \sf E_{(net)}=\sqrt{(200000)^2+ (100000)^2}}$

$\footnotesize{\longrightarrow\: \sf E_{(net)}=\sqrt{ 4 \times 10^{10}+ 1\times 10^{10} }}$

$\footnotesize{\longrightarrow\: \sf E_{(net)}=10^5\sqrt{4 + 1}}$

$\footnotesize{\longrightarrow\: \underline{ \underline{\sf E_{(net)}=\sqrt{5} \times10^5 \:N/C }}}$

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