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When we charge an conductor by induction and grounding, we first bring a negative charge to the conductor. As a result the mobile electrons of the conductor get repelled and stay far from the negative charge. After we ground that conductor and then those repelled electrons go to the earth.

  • But why? There doesn't seem to be any potential difference.

  • So why do those electrons leave the conductor without any reason?

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In your case there will be a negative potential on the conductor and the potential on earth is zero. When you ground it there will be a potential difference, so the electrons will move until the potential difference becomes zero ,i.e. electron will move to ground. Then finally the potential on the conductor will also be zero.

Since, a neutral metal ball have no potential of its own , it will have -ve potential due to external -ve charge.

enter image description here The positive charge induced on one side of sphere will cancel negative charge on other side. Configuration of charges is such that no charge is induced at centre of sphere.

Now potential at centre will only be due to outer negative charge and potential due to two sides of sphere cancel each other exactly.

Potential at centre = potential at any point on sphere = $-kq/r$

where $r$ is distance from center of sphere w.r.t. RHS.

Edit : Another case: If inducing charge is not far.

Consider your -ve charge at distance of $r$ from centre of sphere of radius $R$.

Now, Potential at point near to sphere will be $\frac{-kq}{r-R}$

Potential at far point will be $-kq/(r+R)$

Due to $PD =\frac{kq(2R)}{r^2 - R^2}$ electron move from near side$N$ to far side $F$.

N has positive charge although -ve potential and F has -ve charge .

Now, potential at far side had become more -ve and at near side less -ve.

As +ve charge is induced at $N$ , consider a charge $+q'$ at distance $\frac{r}{R}$ from $N$ which had induced this +ve charge.

Note: there is no such charge q' in real, just let it.

Since net charge on sphere is 0. An equal $-q'$ charge at distance $R/r$ from F will also be formed which had induced -ve charge at F.

Now, due to PD electrons in sphere move,so that, potential at $F$ and $N$ are same.

$$V_{N} = \frac{-kq}{r-R} + \dfrac{kq'}{\frac{R}{r}}+\dfrac{-kq'}{2R-\frac{R}{r}}$$

$$V_{F}=\frac{-kq}{r+R}+\dfrac{-kq'}{\frac{R}{r}}+ \frac{kq'}{2R - \frac{R}{r}}$$

When potential at point N and F will be equal $k\frac{-q}{r-R} + k\dfrac{q'}{\frac{R}{r}} + \frac{-kq'}{2R - \frac{R}{r}} = \frac{-kq}{r+R} + \dfrac{-kq'}{\frac{R}{r}} + \dfrac{kq'}{2R - \frac{R}{r}}$

On solving , $q' = -\frac{qR^2(2r-1)}{2r(r^2- R^2)(1-r)}$

So, potential at N $V_{N}= \frac{-kq}{r-R} + -\frac{kqR^2(2r-1)r}{2r(r^2- R^2)(1-r)R} -\frac{kqrR^2(2r-1)}{2r(r^2- R^2)(1-r)R(2r-1)}$

On solving $V_{N} = -k\frac{qr}{r^2-R^2}$ Which is potential at any point on sphere.

When $r>>R$

$V = \frac{-kq}{r}$

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  • $\begingroup$ ok but how we'll say that mathematically that there is a negative potential on that conductor due to the external charge. $\endgroup$ – ffahim Jan 20 '16 at 14:59
  • $\begingroup$ ok sir tell why u r saying outer POSITIVE CHARGE and how value of Vn is equal to V of any point of the sphere. $\endgroup$ – ffahim Jan 23 '16 at 13:56
  • $\begingroup$ If Vn is different, charge will flow in conductor and make voltage equal again. Here while inducing charge I considered that flow of charge due to unequal potentials. $\endgroup$ – Anubhav Goel Jan 24 '16 at 15:02
  • $\begingroup$ There would be no net potential due to sphere by itself. Potential of metalic sphere can only be due to electric field of outer positive charge. $\endgroup$ – Anubhav Goel Jan 24 '16 at 15:10
  • $\begingroup$ but here inducing charge or outer charge is not positive sir. $\endgroup$ – ffahim Jan 24 '16 at 15:33
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First there will be a potential on the conductor and the potential on earth is zero. when we ground it there will be a potential difference, so the electrons will move until the potential difference becomes zero. Then finally the potential on the conductor will also be zero.

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  • $\begingroup$ ok then what would be the voltage of that conductor.is it positive or negative? $\endgroup$ – ffahim Jan 19 '16 at 13:40
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    $\begingroup$ As we bought a negative charge near to the conductor there will be a negative voltage on conductor. when we ground it the electron will move from low voltage(conductor) to high voltage(earth). $\endgroup$ – Selvaratnam Lavinan Jan 19 '16 at 19:04
  • $\begingroup$ but why there is negative voltage cause the conductor is still neutral(before grounding) why there is voltage difference pls clear me. $\endgroup$ – ffahim Jan 20 '16 at 4:35
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    $\begingroup$ when you bring a negative charge the negative charge will create an electric field. therefore there will be a voltage in every point all over the field. so even though the conductor is neutral there will be a voltage because of the negative charge which is near to it. $\endgroup$ – Selvaratnam Lavinan Jan 20 '16 at 6:23
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    $\begingroup$ voltage on every point of conductor will vary because of the different distance. every electron will feel a different voltage difference and will move according to that. $\endgroup$ – Selvaratnam Lavinan Jan 21 '16 at 4:40

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