Electrostatic Potential and Capacitance

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If there is a potential then E? 0 electric field comes into existence, which is given by as E=-dV/dr
It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. Hence, the entire volume inside must be equipotential.

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As the electric field and force both are conservative so work done in conservative body as the body is closed is zero.

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No, the potential function have no free space because when there is a free space there will be a leakage of potential at that point.

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It is only possible when their sizes are different . when their sizes are different there capacity would change. So there is potential difference btween them.

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The force on a charge particle in electric field F = qE

The free electrons (negative charge) experience electrostatic force in a direction opposite to the direction of electric field.

The direction of electric field is always from higher potential to lower. Hence direction of travel of electrons is from lower potential to region of higher potential.

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Surface charge density is inversely proportional to radius of the sphere so larger sphere has less charge density and vice versa.

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Suppose the charge is slightly shifted to left so net potential energy will be

U= $\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\{\frac{-{\mathit{q}}^2}{\left(\mathit{d}-\mathit{x}\right)}+\frac{-{\mathit{q}}^2}{(\mathit{d}+\mathit{x})}\}$

U= $\frac{-{\mathit{q}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{d}}{({\mathit{d}}^2-{\mathit{x}}^2)}$

$\frac{\mathit{d}\mathit{U}}{\mathit{d}\mathit{x}}$ = $\frac{-{\mathit{q}}^{2}2\mathit{d}}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{x}}{{({\mathit{d}}^2-{\mathit{x}}^2)}^2}$

System will be in equilibrium if this energy is zero

But when we double differentiate it

But when x<0 this value is negative so less than 0

This shows an unstable equilibrium

It follows the parabolic graph

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Let any three point x,y,z  seprated by a distance 2d

Then potential due to two charges

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}+\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$ =0

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}$ = $\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$

X²+Y²+z²+ $\frac{\left(\frac{q1}{q2}\right)+1}{\left(\frac{q1}{q2}\right)-1}$ (2zd)+d²-0

this is equation of sphere .

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$\sigma$ =charge/Area

Q= $\sigma$ (2 $\pi rdr$ )

V=KdQ/ $\surd (r$ ²+x²)=K $\sigma$ (2 $\pi rdr$ )/ $\surd (r$ ²+x²)

After integration we got

V= $k\frac{Q}{2\pi a}$ ₂ ( $\sqrt[]{x^2+a^2}$ )-x

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When initially K₁ is closed and K₂ is open, the capacitors C₁ and C₂ connected in series with battery

Q=CV=C (C₁C₂/C₁+C₂) = 6 $*$ 3/6+3=18 $\mu$ C

Now K₁ was opened and K₂ was closed, the battery and capacitor C, are disconnected from the circuit .

Here charge being shared by both capacitor so a common potential will develop.

V=C₁V₁+C₂V₂/C₁+C₂ = 18/3+3=3V

Q₂=3CV=3 $*$ 3=9μC

Q₃=3CV=3 $*$ 3=9μC