JEE Main & Advanced Physics Electrostatics & Capacitance Neutral Point and Zero Potential

A neutral point is a point where resultant electrical field is zero.

(1) Neutral point Due to a system of two like point charge : For  this case neutral point is obtained at an internal point along the line joining two like charges.

If N is the neutral point at a distance \[{{x}_{1}}\]  from \[{{Q}_{1}}\] and at a distance \[{{x}_{2}}\left( =x-{{x}_{1}} \right)\] from \[{{Q}_{2}}\] then

At N       |E.F. due to \[{{\mathbf{Q}}_{\mathbf{1}}}\]| = |E.F. due to \[{{\mathbf{Q}}_{\mathbf{2}}}\]|

i.e.,  \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}}{x_{1}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{2}}}{x_{2}^{2}}\]\[\Rightarrow \]\[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{2}}\]

Short Trick : \[{{x}_{1}}=\frac{x}{\sqrt{{{Q}_{2}}/{{Q}_{1}}}+1}\] and \[{{x}_{2}}=\frac{x}{\sqrt{{{Q}_{1}}/{{Q}_{2}}}+1}\]

(2) Neutral point due to a system of two unlike point charge : For this condition neutral point lies at an external point along the line joining two unlike charges. Suppose two unlike charge \[{{Q}_{1}}\] and \[{{Q}_{2}}\] separated by a distance \[x\] from each other.

Here neutral point lies outside the line joining two unlike charges and also it lies nearer to charge which is smaller in magnitude.

If \[\left| \left. \,{{Q}_{\mathbf{1}}} \right| \right.<\left| \,\left. {{Q}_{\mathbf{2}}} \right| \right.\] then neutral point will be obtained on the side of \[{{Q}_{1}}\], suppose it is at a distance \[l\] from \[C=\frac{Q}{V}\]

so \[l=\frac{x}{\left( \sqrt{{{Q}_{\mathbf{2}}}/{{Q}_{\mathbf{1}}}}-\mathbf{1} \right)}\]

(3) Zero potential due to a system of two point charge

(i) If both charges are like then resultant potential is not zero at any finite point.

(ii) If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed curve.

(iii) Along the line joining the two charge, two such points exist, one lies inside and one lies outside the charges on the line joining the charges. Both the above points lie nearer the smaller charge.

For internal point

(It is assumed that \[|{{Q}_{1}}\,|\,<\,|{{Q}_{2}}|\]).

At P,   \[\frac{{{Q}_{1}}}{{{x}_{1}}}=\frac{{{Q}_{2}}}{\left( x-{{x}_{1}} \right)}\]

\[\Rightarrow \] \[{{x}_{1}}=\frac{x}{\left( {{Q}_{2}}/{{Q}_{\mathbf{1}}}+1 \right)}\]

For External point

At P,  \[\frac{{{Q}_{1}}}{{{x}_{1}}}=\frac{{{Q}_{2}}}{\left( x+{{x}_{1}} \right)}\]

\[\Rightarrow \] \[{{x}_{1}}=\frac{x}{\left( {{Q}_{2}}/{{Q}_{\mathbf{1}}}-1 \right)}\]