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Two charged spheres of radius R_{1}  and R_{2}   respectively are charged and joined by a wire. The ratio of electric field of the spheres is

  • Option 1)

    \frac{R_{1}}{R_{2}}
     

     

     

  • Option 2)

    \frac{R_{2}}{R_{1}}

  • Option 3)

    \frac{R_{1}^{2}}{R_{2}^{2}}

  • Option 4)

    \frac{R_{2}^{2}}{R_{1}^{2}}

 

Answers (1)

best_answer

As we learned

 

Electric field and potential Due to various charge Distribution i) Point Charge -

Electric field and Potential at Point P due to a Point charge Q.

E= \frac{kQ}{r^{2}}    ,     V= \frac{kQ}{r}

- wherein

 

After connection their potential becomes equal i.e., k\cdot \frac{Q_{1}}{R_{2}}=\frac{k\cdot Q_{2}}{R_{2}};\rightarrow \rightarrow \Rightarrow \frac{Q_{1}}{Q_{2}}=\frac{R_{1}}{R_{2}}

Ratio of electric field \frac{E_{1}}{E_{2}}=\frac{Q_{1}}{Q_{2}}\times \left ( \frac{R_{2}}{R_{1}} \right )^{2}=\frac{R_{2}}{R_{1}}.     


Option 1)

\frac{R_{1}}{R_{2}}
 

 

 

Option 2)

\frac{R_{2}}{R_{1}}

Option 3)

\frac{R_{1}^{2}}{R_{2}^{2}}

Option 4)

\frac{R_{2}^{2}}{R_{1}^{2}}

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