Question

Three point charges of -20√2 nC , +50 nC and - 20√2 nC are placed at the corners A,B and C respectively of a square ABCD of side √2 m. Calculate the resultant electric intensity at the (i) fourth corner D (ii) the centre O of the square.

Answer 1

Solution :

⏭ Given:

✏ Three point charges of -20√2nC, +50nC and -20√2nC are placed at the corners A,B and C respectively of a square ABCD of side √2 m.

⏭ To Find:

✏ Resultant electric field intensity at

• Fourth corner D
• The centre O of the square

⏭ Formula:

✏ Formula of electric field at distance 'r' from a point charge 'q' is given by

$\bigstar \: \boxed{ \tt{ \large{\pink{E = \frac{kq}{ {r}^{2}}}}}} \: \bigstar$

⏭ Calculation:

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• at point D

$\circ \sf \: \red{due \: to \: charge \: A} \\ \\ \implies \sf \: \vec{E}_A = \frac{k(20 \sqrt{2)} }{ { (\sqrt{2}) }^{2} } \\ \\ \implies \sf \: \vec{ E}_A = \frac{20k}{ \sqrt{2} } \: (towards \: DA) \\ \\ \circ \sf \: \red{due \: to \: charge \: B} \\ \\ \implies \sf \: \vec{E}_B = \frac{k(50)}{ {(2)}^{2} } \\ \\ \implies \sf \: \vec{E}_B = 12.5k \: (towards \: BD) \\ \\ \circ \sf \: \red{due \: to \: charge \: C} \\ \\ \implies \sf \: \vec{E}_C = \frac{k(20 \sqrt{2)} }{ {( \sqrt{2} )}^{2} } \\ \\ \implies \sf \: \vec{ E}_C = \frac{20k}{ \sqrt{2} } \: (towards \: DC) \\ \\ \dag \sf \: resultant \: of \: \vec{E}_A \: and \: \vec{E}_C \: ( \vec{E}_A = \vec{E}_C = \vec{E})\\ \\ \mapsto \sf \: \vec{E}_r = \sqrt{ { \vec{E}}^{2} + { \vec{E}}^{2} } \\ \\ \mapsto \sf \: \vec{E}_r = \sqrt{2} \times \vec{ E} \: (towards \: DB)\\ \\ \dag \sf \: Net \: field \: at \: D \: point \\ \\ \leadsto \sf \: \vec{E}_D = \sqrt{2} \vec{E} - \vec{E}_B = \sqrt{2} { \huge{( }}\dfrac{20k}{ \sqrt{2} }{ \huge{ ) }}- 12.5k \\ \\ \leadsto \sf \: \vec{ E}_D = 20k - 12.5k = 7.5 \times 9 \times {10}^{9} \times {10}^{ - 9} \\ \\ \leadsto \sf \: \boxed{ \tt{ \blue{E_D = 67.5 \: N {C}^{ - 1} \: (towards \: DB) }}}$

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• at point O

$\implies \sf \: At \: point \: O \: field \: due \: to \: is \:A \: balanced\\ \sf\: by \: field \: due \: to \: C \\ \\ \therefore \sf \: \vec{E}_O = \vec{E}_B = \frac{k(50 \times {10}^{ - 9}) }{ {(1)}^{2} } \\ \\ \implies \sf \: \vec{E}_O = 9 \times {10}^{9} \times 50 \times {10}^{ - 9} \\ \\ \implies \boxed{ \tt{ \orange{E_O = 450 \: N {C}^{ - 1} \: towards \: BO}}}$

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