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CBSE Class 12-science Answered

Three charges \( 1 \mathrm{C},-1 \mathrm{C} \), and \( 2 \mathrm{C} \) are kept on \( X \) axis \( -5 \mathrm{~cm}, x=0 \mathrm{~cm} \) amd \( x=10 \mathrm{~cm} \) respectively find net force one each of them
Asked by marshelojigas | 28 Jun, 2022, 06:46: AM
Expert Answer
I assume the question is as given below

" Three charges  1 C, -1 C , and  2 C are kept on  X  axis  at  x = -5 cm,
x=0 cm  and  x=10 cm  respectively find net force one each of them "
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Force F between two charges q1 and q2 that are separated by a distance d is given as
 
begin mathsize 14px style F space equals space K space cross times fraction numerator q subscript 1 cross times q subscript 2 over denominator d squared end fraction end style
where K = 1/(4πεo ) = 9 × 109 N m2 C-2 is Coulombs constant and εo is permittivity of free space.
 
In this given charge configuration, there are three types of forces .
 
F1 is attractive force between charges 1 C and -1 C ;  begin mathsize 14px style open vertical bar F subscript 1 close vertical bar space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 1 space cross times 1 over denominator 0.05 cross times 0.05 end fraction equals space 3.600 space cross times 10 to the power of 12 space N end style
 
F2 is attractive force between charges -1 C and 2 C ;  begin mathsize 14px style open vertical bar F subscript 2 close vertical bar space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 1 space cross times 2 over denominator 0.1 cross times 0.1 end fraction equals space 1.800 space cross times 10 to the power of 12 space N end style
F3 is repulsive  force between charges 1 C and 2 C ;  begin mathsize 14px style open vertical bar F subscript 3 close vertical bar space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 1 space cross times 2 over denominator 0.15 cross times 0.15 end fraction equals space 8.00 space cross times 10 to the power of 11 space N end style
Net force on 1 C charge = F1 - F3 = ( 3.600 -0.800) × 1012 N = 2.800 × 1012 N
 
Net force on -1 C charge = F1 - F2 = ( 3.600 -1.800) × 1012 N = 1.800 × 1012 N
 
Net force on 2 C charge = F2 - F3 = ( 1.800 -0.800) × 1012 N = 1.000 × 1012 N
Answered by Thiyagarajan K | 28 Jun, 2022, 08:27: AM

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