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  • I need help with Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulated string of equal length L meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity

Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulated string of equal length L meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity (state of weight less ness). Then the angle between the string and tension in the string is

  • Option 1)

    \\*180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{(2L)^{2}}\; \; \;

  • Option 2)

    \; 90^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{L^{2}}\; \;

  • Option 3)

    \; 180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{2L^{2}}\; \;

  • Option 4)

    \; 180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{QL}{4L^{2}}

 

Answers (1)

As we learned

Coulombic force -

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 In case to weight less ness following situation arises.  So

angle\; \; \theta =180^{\circ}\; and\; force\; F=\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{(2L)^{2}}                                

 


Option 1)

\\*180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{(2L)^{2}}\; \; \;

Option 2)

\; 90^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{L^{2}}\; \;

Option 3)

\; 180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{Q^{2}}{2L^{2}}\; \;

Option 4)

\; 180^{\circ},\frac{1}{4\pi \varepsilon _{0}}.\frac{QL}{4L^{2}}

Posted by

Vakul

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