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 th

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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 hig

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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 follo

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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 $\times$ 3/6+3=18 $\mu$ C

Now K₁ was opened and K₂ was closed, the battery and capacitor C,