3point charge q, -2q, and 4q are kept at 3 vertices of a equilateral triangle find net force on each of them

Asked by marshelojigas | 27th Jun, 2022, 06:03: AM

Expert Answer:

 
 
Figure shows the charge configuration on an equilateral triangle of side a as given in the question .
 
Type of force between charges,  whetherforce is  attractive or repulsive,  is decided depends on sign of charges .
 
Force FA on a charge q at vertex A is determined as follows .

begin mathsize 14px style F subscript A equals space K space cross times fraction numerator 4 q squared over denominator a squared end fraction cross times left parenthesis space cos 30 space i with hat on top space minus space c o 60 space j with hat on top space right parenthesis plus space K cross times fraction numerator 2 space q squared over denominator a squared end fraction cross times left parenthesis negative cos 60 space space i with hat on top space space minus space cos 30 space space j with hat on top space right parenthesis end style
where K = 1/ ( 4πεo ) is Coulomb's constant  , a is side of triangle ,
begin mathsize 14px style i with hat on top end style is unit vector along +x axis direction and begin mathsize 14px style j with hat on top space end style is +y axis direction
begin mathsize 14px style F subscript A equals space K space cross times fraction numerator 2 q squared over denominator a squared end fraction cross times open square brackets left parenthesis 2 cos 30 minus cos 60 right parenthesis space space i with hat on top space space minus space left parenthesis 2 cos 60 space plus space cos 30 space right parenthesis j with hat on top close square brackets end style
begin mathsize 14px style F subscript A equals space K space cross times q squared over a squared cross times open square brackets left parenthesis 2 square root of 3 minus 1 right parenthesis space space i with hat on top space space minus space left parenthesis 2 square root of 3 space plus space 1 space right parenthesis j with hat on top close square brackets end style
Magnitude of Fis given as
 
begin mathsize 14px style open vertical bar F subscript A close vertical bar space equals space K space cross times q squared over a squared cross times square root of left parenthesis 2 square root of 3 space minus space 1 space right parenthesis squared plus left parenthesis 2 square root of 3 space plus space 1 space right parenthesis squared end root space equals space K space cross times q squared over a squared cross times square root of 26 end style
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Force FB on a charge -2q at vertex B is determined as follows .

begin mathsize 14px style F subscript B space equals space space space K cross times fraction numerator 8 q squared over denominator a squared end fraction space i with hat on top space plus K cross times fraction numerator 2 q squared over denominator a squared end fraction left parenthesis cos 60 space space i with hat on top space plus space cos 30 space j with hat on top space right parenthesis end style
 
 
begin mathsize 14px style F subscript B space equals space space space K cross times q squared over a squared left parenthesis 9 space i with hat on top space plus space square root of 3 space space j with hat on top space right parenthesis end style
Magnitude of Fis given as
 
begin mathsize 14px style open vertical bar F subscript B close vertical bar space equals space K space cross times q squared over a squared cross times space square root of 9 squared plus 3 squared end root space equals K space cross times q squared over a squared cross times 2 square root of 21 end style
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Force FC on a charge 4q at vertex C is determined as follows .

begin mathsize 14px style F subscript C space equals space K space cross times fraction numerator 8 q squared over denominator a squared end fraction left parenthesis negative i with hat on top space right parenthesis space plus space K space cross times fraction numerator 4 q squared over denominator a squared end fraction space cos 60 space i with hat on top space minus space space K space cross times fraction numerator 4 q squared over denominator a squared end fraction space cos 30 space j with hat on top end style
 
 
 
begin mathsize 14px style F subscript C space equals space K space cross times fraction numerator 2 q squared over denominator a squared end fraction left parenthesis negative 3 i with hat on top space space minus space square root of 3 space space j with hat on top right parenthesis space space space end style
Magnitude of FC is given as
 
begin mathsize 14px style space open vertical bar F subscript C close vertical bar equals space K space cross times fraction numerator 2 q squared over denominator a squared end fraction left parenthesis space square root of 3 squared plus 3 end root space right parenthesis space equals space K space cross times fraction numerator 4 q squared over denominator a squared end fraction cross times space square root of 3 space space space end style

Answered by Thiyagarajan K | 27th Jun, 2022, 08:42: AM

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