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CBSE Class 12-science Answered

Two equal small spheres each weighing 1g , hang by two equal silk threads from the same point. The spheres are charged when in contact and come to rest with their centres 2cm apart and 20cm vertically below the point of support. Find the charge on each sphere. 
Asked by ishantkundu31 | 14 Jun, 2021, 05:00: PM
answered-by-expert Expert Answer
Figure shows forces acting on charged sphere. Weight mg is acting downward.  Repulsive electrostatic force is shown as Fe .
 
Tension force T acting along string is resolved as ( T cosθ ) and ( T sinθ ) . 

(Tcosθ) component balances weight mg.   (T sinθ) component balances Electrostatic force Fe . 
 
T cosθ = mg ...................... (1)

T sinθ  = Fe ........................(2)
 
If we eliminate tension T in above equations, we get ,  Fe = m g tanθ .......................(3)
 
Eqn.(3) is written as
 
K × ( q2 / d2 ) = ( m g tanθ ) ........................... (4)
 
where K = 1/( 4 π εo ) = 9 × 109 N m2 C-2 Coulomb's constant  and d = 2 cm  is the distance between charged spheres.
 
m is mass of sphere , g is acceleration due to gravity and θ is the angle made by the string with vertical
 
From eqn.(4) , we get 
 
begin mathsize 14px style q space equals space square root of open parentheses fraction numerator m space g space tan theta space d squared over denominator K end fraction close parentheses end root space space end style
 
begin mathsize 14px style q space equals space square root of fraction numerator 10 to the power of negative 3 end exponent cross times 9.8 cross times left parenthesis 1 divided by 20 right parenthesis cross times 4 cross times 10 to the power of negative 4 end exponent over denominator 9 cross times 10 to the power of 9 end fraction end root space equals 4.667 cross times 10 to the power of negative 9 space end exponent C space equals space 4.667 space n C end style
Answered by Thiyagarajan K | 14 Jun, 2021, 09:41: PM

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