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

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May 2021 1 55 Report

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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TPS

Verified answer

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

raoatchut191 b)(q)sq(2rt2+1)/4 pie epsolonnot (l)sq

raoatchut191 c)(q)sq(2rt2-1/4pie epsolonnot (l)sq

raoatchut191 d)(q)sq(2rt2+1/4pie epsolonnot 2 (l)sq

raoatchut191 these are given options

raoatchut191 are u able to understand?

TPS yess. answer is a. ignore the minus sign from q as it just shows the direction of force. your option contains only magnitude.

raoatchut191 thnq very much

TPS Ok now??

raoatchut191 yes

raoatchut191 willu answer another one

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