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The potential (in volts) of a charge distribution is given by

 V(z) = 30 - 5z^{2} for\left | z \right | \leqslant 1m 

V(z) = 35 - 10\left | z \right | for \left | z \right | \geqslant 1m

V(z)  does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of  \epsilon _{0} ) which is spread over a certain region, then choose the correct statement.

  • Option 1)

    \rho _{0} = 10 \, \epsilon _{0}\, for \left | z \right |\leqslant 1 m\, and\, \rho _{0}\, = 0\, elsewhere

  • Option 2)

    \rho _{0} = 20\, \varepsilon _{0}\, in \, the\, entire\, region

  • Option 3)

    \rho _{0} =40 \, \epsilon _{0} \, in \, the \, entire\, region

  • Option 4)

    \rho _{0} = 20\, \epsilon _{0}\, for\, \left | z \right |\leqslant 1m\: and\, \rho _{0} = 0\: elsewhere

 

Answers (1)

best_answer

As we learnt in

Electric Potential due to Continious charge distribution -

\dpi{100} V=\int dV=\int \frac{dq}{4\pi \varepsilon _{0}r}

-

 

 

Given\ V(z)=\left \{ 30-5z^{2}\left. \right \}\: for |z|\leq 1m

                     =\lef \{ 35-10|z| \left. \right \}for |z|\geq 1m

Now E(z)=\, -\frac{dV}{dz}= 10z \, \, for |z| \leq 1m = 10 for |z| \geq 1m

\therefore The source is an infinite non conducting thick plate of thickness  2m

\thereforeE= \frac{q}{2AE_{0}}= \frac{q}{Af}= \rho _0= \frac{2E}{4}\, E_{0}= \frac{2\times 10}{2}\, E_{0}

\therefore E= 10E_{0}


Option 1)

\rho _{0} = 10 \, \epsilon _{0}\, for \left | z \right |\leqslant 1 m\, and\, \rho _{0}\, = 0\, elsewhere

Correct Option

Option 2)

\rho _{0} = 20\, \varepsilon _{0}\, in \, the\, entire\, region

Incorrect Option

Option 3)

\rho _{0} =40 \, \epsilon _{0} \, in \, the \, entire\, region

Incorrect Option

Option 4)

\rho _{0} = 20\, \epsilon _{0}\, for\, \left | z \right |\leqslant 1m\: and\, \rho _{0} = 0\: elsewhere

Incorrect Option

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Aadil

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