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Two point charges $+8q$ and $-2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x-$ axis at which the net electric field due to these two point charges is zero is
$8\,L$
$4\,L$
$2\,L$
$\frac {L}{4}$
Solution
The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude.
Suppose $\mathrm{E.F.}$ is zero at $\mathrm{P}$ as shown.
Hence at $\mathrm{P} ; \mathrm{k}=\frac{8 \mathrm{q}}{(\mathrm{L}+l)^{2}}=\frac{\mathrm{k} \cdot(2 \mathrm{q})}{l^{2}} \Rightarrow l=\mathrm{L}$
Similar Questions
In steady state heat conduction, the equations that determine the heat current $j ( r )$ [heat flowing per unit time per unit area] and temperature $T( r )$ in space are exactly the same as those governing the electric field $E ( r )$ and electrostatic potential $V( r )$ with the equivalence given in the table below.
| Heat flow | Electrostatics |
| $T( r )$ | $V( r )$ |
| $j ( r )$ | $E ( r )$ |
We exploit this equivalence to predict the rate $Q$ of total heat flowing by conduction from the surfaces of spheres of varying radii, all maintained at the same temperature. If $\dot{Q} \propto R^{n}$, where $R$ is the radius, then the value of $n$ is
