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  • Help me please, A thin spherical conducting shell of radius R has a charge q . Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is

A thin spherical conducting shell of radius  R has a charge q . Another charge  Q   is placed at the centre of the shell. The electrostatic potential at a point  P at a distance  R/2 from the centre of the shell is

  • Option 1)

    \frac{2Q}{4\pi \varepsilon _{0}R}

  • Option 2)

    \frac{2Q}{4\pi \varepsilon _{0}R}-\frac{2q}{4\pi \varepsilon _{0}R}

  • Option 3)

    \frac{2Q}{4\pi \varepsilon _{0}R}+\frac{q}{4\pi \varepsilon _{0}R}

  • Option 4)

    \frac{\left ( q+Q \right )}{4\pi \varepsilon _{0}}\frac{2}{R}

 

Answers (2)

best_answer

As we learnt in

If P at the surface r = R -

E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}       V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}

\dpi{100} E_{s}=\frac{\rho R}{3 \epsilon _{0}}               V_{s}=\frac{\rho R^{2}}{3 \epsilon _{0}}

-

 

 Potential at any internal point of charged shell         = \frac{q}{4\pi \varepsilon _{0}R}

potential at  P  due to  Q   at centre       = \frac{1}{4\pi \varepsilon _{0}}\frac{2Q}{R}

Total potential point  = \frac{q}{4\pi \varepsilon _{0}R}+\frac{2Q}{4\pi \varepsilon _{0}R}= \frac{1}{4\pi \varepsilon _{0}R}\left (q+2Q \right )

 


Option 1)

\frac{2Q}{4\pi \varepsilon _{0}R}

Incorrect

Option 2)

\frac{2Q}{4\pi \varepsilon _{0}R}-\frac{2q}{4\pi \varepsilon _{0}R}

Incorrect

Option 3)

\frac{2Q}{4\pi \varepsilon _{0}R}+\frac{q}{4\pi \varepsilon _{0}R}

Correct

Option 4)

\frac{\left ( q+Q \right )}{4\pi \varepsilon _{0}}\frac{2}{R}

Incorrect

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Plabita

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