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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-----------------

Shreya D Nanda
74 Points
3 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, $\dpi{80} r_{1}=r$ and $\dpi{80} F_{1} \propto 1/[r_{1}]^2$ .$\dpi{80} \Rightarrow F_{1} \propto 1/r^2$
In the next case, as given $\dpi{80} r_{2}=3r$ and $\dpi{80} F_{2} \propto 1/[r_{2}]^2$ . $\dpi{80} \Rightarrow F_{2} \propto 1/[3r]^2$
Taking ratio, $\dpi{80} F_{1}/F_{2}= r_2^2 /r_1 ^2= [3r]^2/r^2$
$\dpi{80} \Rightarrow F_{1}/F_{2}= 9$                   but force $\dpi{80} F_{1} =18 N$
$\dpi{80} \Rightarrow F_{2}=18/9 = 2 N$
For electric field, $\dpi{80} E=F/q$
$\dpi{80} \therefore E_{1}/E_{2} = F_{1}/F_{2} =9$
or  $\dpi{80} E_{2}=1/9 E_{1}$    or   $\dpi{80} E_{2}= F_{2}/q =2/q NC^{-1}$
If q=1 C, then $\dpi{80} E_{2}=2 NC^{-1}$

hope it helps