Sir answer for the question which I have attached as a image
Sir answer for the question which I have attached as a image
gokjaghem14 , 5 Years ago
Grade
Electrostatics> Sir answer for the question which I have ...
2 AnswersParth
Last Activity: 5 Years ago
See in this case
Work done = change in potential energy + K.E. here KE = 0
assuming charge At a = 1
so U1 – U2
U1 intial Potential Energy
U1 = k*(q^2)/a + 2*k*q*1/a
for final configuration
U2 = k*(q^2)/a + 2*k*(q)/(a/sqrt(3))
so |U1- U2| =
(2*k*q*(3^(1/2) - 1))/a
So option B
pls Approve if it helps
Ayush_Deep
Last Activity: 5 Years ago
The correct answer will be option B .
Work done in moving a charge particle is equal to change in the potential energy of particle = final PE - initial PE
The initial potential energy is 2kq/a
a is side of equilateral triangle.
Final potential energy is 2kq/(a/3½)=2√3kq/a
Work done = kq(2√3-2)/a= 2kq(√3-1)/a
Option B .
Thanks
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