At-each-of-three-corners-of-a-square-of-side-2cm-is-placed-a-point-charge-of-magnitude-3-C-What-will-be-the-magnitude-and-direction-of-the-resultant-force-on-a-point-charge-1-C-if-it-were-placed-

Question Number 46374 by Umar last updated on 24/Oct/18

A t e a c h o f t h r e e c o r n e r s o f a s q u a r e

o f s i d e 2 c m i s p l a c e d a p o i n t c h a r g e

o f m a g n i t u d e 3 μ C .

W h a t w i l l b e t h e m a g n i t u d e a n d

d i r e c t i o n o f t h e r e s u l t a n t f o r c e o n a

p o i n t c h a r g e - 1 μ C i f i t w e r e p l a c e d

(a) a t t h e c e n t r e o f t h e s q u a r e ?

(b) a t t h e v a c a n t c o r n e r o f t h e s q r e ?

Commented by Umar last updated on 24/Oct/18

p l e a s e h e l p

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Oct/18

l e t S q u a r e A B C D \dots

a t A (0, 0) c h a r g e 3 μ C

a t B (2, 0) c h a r g e 3 μ C

a t C (2, 2) n o c h a r g e

a t D (0, 2) c h a r g e 3 μ C

a t E (1, 1) t h e c e n t r e c h a r g e (- 1 μ C)

f o r c e o n E b y A i s F_{E A} a t t r a c t i v e

f o r c e o n E b y B i s F_{E B a t t r a c t i v e}

f o r c e o n E b y D i s F_{E D a t t r a c t i v e}

b u t f o r c e F_{E B} i s e w u a l i n m a g n e t u d e w i t h

F_{E D} b u t o p p o s i t e i n d i r e c t i o n s o c a n c e l l e a c h o t h e r

s o n e t f o r c e o n E i s F_{E A}

{\vec{F}}_{E A} = \frac{1}{4 π ϵ_{0}} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(\sqrt{2} \times 10^{- 2})}^{2}} d i r e c t i o n E \to A

= 9 \times 10^{9} \times \frac{3 \times 10^{- 12}}{2 \times 10^{- 4}} = 135 N

s e c o n d q u e s t i o n \dots

w h e n c h a r g e (- 1 μ C) p l a c e d a t c o r n e r C

f o r c e a r e F_{C B}, F_{C D} a n d F_{C A}

F_{C B} m a g n e t u d e = F_{C D} m a g n e t u d e s a y (F)

F = 9 \times 10^{9} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(2 \times 10^{- 2})}^{2}} N = 6.75 N

R e s u l t a n t f o r c e o f F_{C B} a n d F_{C D} = F_{R} (s a y)

F_{R} = \sqrt{F^{2} + F^{2} + 2 F \times F c o s 90^{0}} = \sqrt{2} F d i r e c t i o n C \to A

F_{R} = \sqrt{2} \times 6.75 N = 1.41 \times 6.75 \approx 9.52 N

n o w F_{C A} = 9 \times 10^{9} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(2 \sqrt{2} \times 10^{- 2})}^{2}} d i r e c t i o n C \to A

F_{C A} = \frac{27}{8} \times 10 = 33.75 N

s o n e t f o r c e = (9.52 + 33.75) N = 43.27 N d i r e c t i o n

C \to A

p l s c h e c k \dots

Commented by Umar last updated on 31/Oct/18

t h a n k y o u s i r

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