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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

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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