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Electric force between two charged spheres is 18 units.If the distance between the centres of the sphere is tripled,the electric field will be-----------------

SOWMYA , 7 Years ago
Grade 12th pass
anser 1 Answers
Shreya D Nanda

Last Activity: 7 Years ago

HI,
Electric force, F=\left (1/4\pi \epsilon _{0})q_{1} q_{2}/r^{2}.. \Rightarrow F\propto 1/r^2
Initially consider distance, r_{1}=r and F_{1} \propto 1/[r_{1}]^2 .\Rightarrow F_{1} \propto 1/r^2
In the next case, as given r_{2}=3r and F_{2} \propto 1/[r_{2}]^2 . \Rightarrow F_{2} \propto 1/[3r]^2
Taking ratio, F_{1}/F_{2}= r_2^2 /r_1 ^2= [3r]^2/r^2
                  \Rightarrow F_{1}/F_{2}= 9                   but force F_{1} =18 N
                  \Rightarrow F_{2}=18/9 = 2 N
For electric field, E=F/q
  \therefore E_{1}/E_{2} = F_{1}/F_{2} =9
 or  E_{2}=1/9 E_{1}    or   E_{2}= F_{2}/q =2/q NC^{-1}
If q=1 C, then E_{2}=2 NC^{-1}
 
hope it helps
 

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