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ABCD is a square of side "l" .the force at b due to the remaining charges is ..........

i need solution
the diagram is given below
pls answer faaaaaaaaaast




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Force acts on charge at B due to charges at A,C and D.

Force between two charges =  \frac{1}{4 \pi e}  \frac{q_{1}q_{2} }{r^{2} } along the radial direction.

Force on charge at B due to charge at A(Fa) =  \frac{1}{4 \pi e} \frac{q×(-q) }{r^{2} }
                                                                 =  \frac{1}{4 \pi e}  \frac{-q^{2} }{I^{2} } (towards A along line AB)

Force on charge at B due to charge at C(Fc) =  \frac{1}{4 \pi e} \frac{q×(-q) }{r^{2} }
                                                                 =  \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} } (towards C along line BC)

Force on charge at D due to charge at A(Fd) =  \frac{1}{4 \pi e } \frac{q×q }{r^{2} }
                                                                 =  \frac{1}{4 \pi e } \frac{q^{2} }{(√2I)^{2} }
                                                                 =  \frac{1}{4 \pi e } \frac{q^{2} }{2I^{2} }  (towards B along the line DB)

Resultant of force acting on B due to A and C (Fac)= √[2×( \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }  )²]
                                                                      = √2 ×  \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }  (towards D along the line DB)

Resultant force (R)= Fac - Fd = √2 ×  \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} } +  \frac{1}{4 \pi e } \frac{q^{2} }{2I^{2} }  

⇒ R = (√2 - 1/2)  \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }

⇒R = ( \frac{(2 \sqrt{2} - 1)}{2} )( \frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} } )  (towards D along the line BD)

⇒|R| =  \frac{(2 \sqrt{2} - 1) q^{2}  }{4 \pi  e * 2 I^{2}}

1 5 1
yess. answer is a. ignore the minus sign from q as it just shows the direction of force. your option contains only magnitude.
thnq very much
Ok now??
willu answer another one
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