Two point charges + Q, and-Q, are placed r distance apart. Obtain
the expression for the amount of work done to place a third charge
Q3 at the midpoint of the line joining the two charges.
(b) At what distance from charge + Q, on the line joining the two
charges (in terms of Q1, Q2 and r) will this work done be zero.
Answers
Answer:
(a)as we know:W=q*V
now work done to place the third charge charge at mid point:
let the distance between two charges +Q and -Q be "r" ,
now their mid point is r/2
therefore,W=q(Q3)*V{(+2Q/4∈r)+(-2Q/4∈r)}
W=Q3*2/4∈r(+Q-Q)
W=Q3*2/4∈r(0)
W=0
(b)we have already found out were W is 0.but still.......
let at a point P at a distance x from +Q ,V is zero hence W is also zero at
distance x because W=q*V
if V =0,then W=0
now,V= (+Q/4∈x)+(-Q/4∈(r-x))
A.T.Q :(+Q/4∈x)+(-Q/4∈(r-x))=0
(1/x-1/(r-x))*Q/4∈=0
(1/x-1/(r-x)=0
1/x=1/(r-x)
x=r-x
2x=r
therefore,x=r/2
hence at r/2 distance from +Q ,W is zero.
Explanation:(a) Here q=Q3 (charge that we want to place )
V=(potential difference due to charge +Q at distance r/2)+(potential difference due to charge +Q at distance r/2)
(b)x=distance between +Q and point P
(r-x)=distance between -Q and point P
The expression for the work done W = (Q₃/2πε₀r)(Q₁ - Q₂)
and the required distance is Q₁r/ (Q₂ + Q₁).
Given:
Two point charges + Q₁, and - Q₂, are placed r distance apart and the third charge is Q₃ at the midpoint of the line joining the two charges.
To find:
(a) Obtain the expression for the amount of work done to place a third charge Q₃
(b) At what distance from charge + Q, on the line joining the two charges (in terms of Q₁, Q₂, and r) will this work done be zero?
Solution:
Let A, B be the point at charges + Q₁, and - Q₂ and C be the point at Q₃
To find the work done find the potential difference between infinity to point C i.e that is given by U - U Hence
Work done, W = Q₃(U - U)
=> W = Q₃(U) [ ∵ U = 0 ]
As we know potential at Q₃ is the potential difference between Q₁ and Q₂
=> W = Q₃ [ Q₁/4πε₀(r/₂) - Q₂/4πε₀(r/₂) ]
=> W = Q₃ [ Q₁/2πε₀r - Q₂/2πε₀r]
=> W = Q₃ (1/2πε₀r)(Q₁ - Q₂)
=> W = (Q₃/2πε₀r)(Q₁ - Q₂)
Hence, The work done W = (Q₃/2πε₀r)(Q₁ - Q₂)
Here, when the potential of point C becomes zero then the work done, will be zero. Let 'x' be the distance of point c from Q₁ and the potential is zero
Hence, the net potential at C
=> K [ Q₁/x - Q₂/(r-x) ] = 0
=> [ Q₁/x - Q₂/(r-x) ] = 0
=> [ Q₁/x - Q₂/(r-x) ] = 0
=> Q₁/x = Q₂/(r-x)
=> Q₁r- Q₁x = Q₂x
=> Q₁r = Q₂x + Q₁x
=> x(Q₂ + Q₁) = Q₁r
=> x = Q₁r/ (Q₂ + Q₁)
The expression for the work done W = (Q₃/2πε₀r)(Q₁ - Q₂)
and the required distance is Q₁r/ (Q₂ + Q₁)
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