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Two charges +\mathrm{q} and +9 \mathrm{q} are separated by a distance of 10 \mathrm{a}. Find the point on the line joining the two charges where electric field is zero.
Physics | Previous Year Question Papers
Two charges +\mathrm{q} and +9 \mathrm{q} are separated by a distance of 10 \mathrm{a}. Find the point on the line joining the two charges where electric field is zero.

Ans: Let p be the point (at x distance from charge +q ) on the line joining the given two charges where the electric field is zero.
We know that the electric field at a point at r distance from any charge a is given by,

    \[E=K \frac{q}{r^{2}}\]

Electric field due to charge +q at point P would be,

    \[E_{1}=K \frac{(+q)}{x^{2}} \ldots \ldots(1)\]

Electric field due to charge +9 \mathrm{q} at point \mathrm{P} would be,

    \[E_{1}=K \frac{(+9 q)}{(10 a-x)^{2}} \ldots \ldots(2)\]

Since the net electric field at point \mathrm{P} is zero,

    \[\begin{array}{l} E_{1}+E_{2}=0 \\ \Rightarrow\left|E_{1}\right|=\left|E_{2}\right| \end{array}\]

From (1) and (2),

    \[\begin{array}{l} \Rightarrow \mathrm{K} \frac{9}{\mathrm{x}^{2}}=\mathrm{K} \frac{9 \mathrm{q}}{(10 \mathrm{a}-\mathrm{x})^{2}} \\ \Rightarrow(10 \mathrm{a}-\mathrm{x})^{2}=9 \mathrm{x}^{2} \\ \Rightarrow 10-\mathrm{x}=3 \mathrm{x} \\ \Rightarrow 10 \mathrm{a}=4 \mathrm{x} \\ \therefore \mathrm{x}=\frac{10}{4} \mathrm{a}=2.5 \mathrm{a} \end{array}\]

Therefore, we found the point on the line joining the given two charges where the net electric field is zero to be at a distance x=2.5 a from charge a and at a distanoe
Where, F is the force on the dipole and x is the perpendicular distance. Where, force F is given by,

    \[\mathrm{F}=\mathrm{q} \mathrm{E} \ldots \ldots(3)\]

From the figure we have,

    \[\begin{array}{l} \sin \theta=\frac{B N}{A B} \\ \Rightarrow B N=A B \sin \theta=2 \mid \sin \theta \end{array}\]

But eN here is the perpendicular distance x, so, equation (2) becomes, \tau=\mathrm{qE} \times 2 \mid \sin \theta=(2 \mathrm{q}) \mathrm{E} \sin \theta
But from (1), \mathrm{P}=2 \mathrm{dq}
Now, we could give the torque on the dipole as,

    \[\therefore \tau=P E \sin \theta=P \times E\]

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