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Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of electric field of the spheres is

  • Option 1)

    a/b

     

     

  • Option 2)

    b/a

  • Option 3)

    a^{2}/b^{2}

  • Option 4)

    b^{2}/a^{2}

 

Answers (1)

best_answer

As we learned

 

Outside the sphere (P lies outside the sphere) -

\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}

V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}=\frac{\sigma R^{2}}{\epsilon _{0}r}

 

- wherein

\sigma - surface charge density.

 

Joined by a wire means they are at the same potential. For same potential \frac{kQ_{1}}{a_{1}}=\frac{kQ_{2}}{a_{2}}\Rightarrow \frac{Q_{1}}{Q_{2}}=\frac{a}{b}

Further, the electric field at the surface of the sphere having radius R and charge Q is \frac{kQ}{R^{2}}

\frac{E_{1}}{E_{2}}=\frac{kQ_{1}/a^{2}}{kQ_{2}/b^{2}} = \frac{Q_{1}}{Q_{2}}\times \frac{b^{2}}{a^{2}}=\frac{b}{a}   


Option 1)

a/b

 

 

Option 2)

b/a

Option 3)

a^{2}/b^{2}

Option 4)

b^{2}/a^{2}

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Plabita

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