At each of three corners of a square of side 2cm is placed a point charge of magnitude 3μC. What will be the magnitude and direction of the resultant force on a point charge −1μC if it were placed (a) at the centre of the square? (b) at the vacant corner of the sqre?


Question and Answers Forum

All Questions Topic List
Electrostatics Questions
Previous in All Question Next in All Question
Previous in Electrostatics Next in Electrostatics
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

	$l e t S q u a r e A B C D . . .$ $a t A (0, 0) c h a r g e 3 μ C$ $a t B (2, 0) c h a r g e 3 μ C$ $a t C (2, 2) n o c h a r g e$ $a t D (0, 2) c h a r g e 3 μ C$ $a t E (1, 1) t h e c e n t r e c h a r g e (- 1 μ C)$ $f o r c e o n E b y A i s F_{E A} a t t r a c t i v e$ $f o r c e o n E b y B i s F_{E B a t t r a c t i v e}$ $f o r c e o n E b y D i s F_{E D a t t r a c t i v e}$ $b u t f o r c e F_{E B} i s e w u a l i n m a g n e t u d e w i t h$ $F_{E D} b u t o p p o s i t e i n d i r e c t i o n s o c a n c e l l e a c h o t h e r$ $s o n e t f o r c e o n E i s F_{E A}$ ${\vec{F}}_{E A} = \frac{1}{4 π ϵ_{0}} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(\sqrt{2} \times 10^{- 2})}^{2}} d i r e c t i o n E \to A$ $= 9 \times 10^{9} \times \frac{3 \times 10^{- 12}}{2 \times 10^{- 4}} = 135 N$ $s e c o n d q u e s t i o n . . .$ $w h e n c h a r g e (- 1 μ C) p l a c e d a t c o r n e r C$ $f o r c e a r e F_{C B}, F_{C D} a n d F_{C A}$ $F_{C B} m a g n e t u d e = F_{C D} m a g n e t u d e s a y (F)$ $F = 9 \times 10^{9} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(2 \times 10^{- 2})}^{2}} N = 6.75 N$ $R e s u l t a n t f o r c e o f F_{C B} a n d F_{C D} = F_{R} (s a y)$ $F_{R} = \sqrt{F^{2} + F^{2} + 2 F \times F c o s 90^{0}} = \sqrt{2} F d i r e c t i o n C \to A$ $F_{R} = \sqrt{2} \times 6.75 N = 1.41 \times 6.75 \approx 9.52 N$ $n o w F_{C A} = 9 \times 10^{9} \times \frac{(1 \times 10^{- 6}) (3 \times 10^{- 6})}{{(2 \sqrt{2} \times 10^{- 2})}^{2}} d i r e c t i o n C \to A$ $F_{C A} = \frac{27}{8} \times 10 = 33.75 N$ $s o n e t f o r c e = (9.52 + 33.75) N = 43.27 N d i r e c t i o n$ $C \to A$ $p l s c h e c k . . .$
	Commented by Umar last updated on 31/Oct/18

	$t h a n k y o u s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum