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Charge Sharing Between Conductors

Charge Sharing Between Conductors :- When two isolated conductors are connected by a conducting wire, charge starts flowing between them until they reach the same electric potential. This happens because charge always moves from higher potential to lower potential to minimize potential difference.

Process of Charge Sharing Between Conductors :

If the conductors are identical (same shape and size), the final charge on both will be equal.
If the conductors have different sizes (different capacitances), the charges distribute in such a way that their potentials become equal, not necessarily their charges.

Let us consider two spherical conductors with different radii, and , carrying charges and respectively. The spheres are placed far apart so that they do not influence each other initially. When they are connected by a conducting wire, charge flows between them until both conductors attain the same electric potential. Let the final charges on the spheres after connection be and respectively.

As after joining they are at same potential, hence :

$\displaystyle \frac{k{{q}_{1}}}{{{R}_{1}}}=\frac{k{{q}_{2}}}{{{R}_{2}}}$

$\displaystyle \Rightarrow \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}$ …..(1)

So final charges will be proportional to their radii.

Also net charge of the system is conserved, hence :

$\displaystyle {{q}_{1}}+{{q}_{2}}={{Q}_{1}}+{{Q}_{2}}$ …..(2)

From equations (1) and (2),

$\displaystyle {{q}_{1}}=\left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right)\left( {{Q}_{1}}+{{Q}_{2}} \right)$ …..(3)

$\displaystyle {{q}_{2}}=\left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\left( {{Q}_{1}}+{{Q}_{2}} \right)$ …..(4)

Ratio of final charges :

$\displaystyle \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{\left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right)\left( {{Q}_{1}}+{{Q}_{2}} \right)}{\left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\left( {{Q}_{1}}+{{Q}_{2}} \right)}$

$\displaystyle \Rightarrow \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}$ …..(5)

Ratio of final surface charge densities :

$\displaystyle \frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{\frac{{{q}_{1}}}{4\pi R_{1}^{2}}}{\frac{{{q}_{2}}}{4\pi R_{2}^{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}$

Using equation (5),

$\displaystyle \frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}$

$\displaystyle \Rightarrow \frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}$ …..(6)

Loss of energy during charge sharing between conductors

During charge sharing, a current flows from the sphere at higher potential to the sphere at lower potential through the wire and some energy is lost in the form of heat. This lost energy will be equal to the loss of electrostatic potential energy. Hence loss of energy (E) in the wire

$\displaystyle E={{U}_{i}}-{{U}_{f}}={{\left[ \frac{kQ_{1}^{2}}{2{{R}_{1}}}+\frac{kQ_{2}^{2}}{2{{R}_{2}}} \right]}_{initial\text{ }self\text{ }energy}}-{{\left[ \frac{kq_{1}^{2}}{2{{R}_{1}}}+\frac{kq_{2}^{2}}{2{{R}_{2}}} \right]}_{final\text{ }self\text{ }energy}}$

Since the spheres are far apart from each other, we have not considered the interaction energy between them.

Example 1.

An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high resistance wire. After a long time is the amount of heat loss?

Solution :

When two conducting spheres of equal radii are connected, the charge gets equally distributed between them. Hence, the heat lost :

$\displaystyle E={{U}_{i}}-{{U}_{f}}={{\left[ \frac{k{{Q}^{2}}}{2R}+0 \right]}_{initial\text{ }self\text{ }energy}}-{{\left[ \frac{k{{\left( {}^{Q}\!\!\diagup\!\!{}_{2}\; \right)}^{2}}}{2R}+\frac{k{{\left( {}^{Q}\!\!\diagup\!\!{}_{2}\; \right)}^{2}}}{2R} \right]}_{final\text{ }self\text{ }energy}}$