Practicing Success
Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis. The charge q at origin in equilibrium will have |
maximum force and minimum potential energy. minimum force & maximum potential energy. maximum force & maximum potential energy. minimum force & minimum potential energy. |
minimum force & minimum potential energy. |
The net force on q at origin is $\vec{F}=\vec{F}_1+\vec{F}_2=\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{r^2} \hat{i}+\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{r^2}(-\hat{i})=\hat{0}$ The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is U = $\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{x}+\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{\left(a^2+x\right)}$ $=\frac{1}{4 \pi \varepsilon_0} . q^2\left[\frac{1}{a-x}+\frac{1}{a+x}\right]$ $\frac{d U}{d x} =\frac{q^2}{4 \pi \varepsilon_0}\left[\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\right]$ For U to be minimum $\frac{d U}{d x}=0, \quad \frac{d^2 U}{d x^2}>0$, $\Rightarrow(a-x)^2=(a+x)^2$ $\Rightarrow a+x= \pm(a-x)$ ⇒ x = 0, because other solution is irrelevent. Thus the charged particle at the origin will have minimum force and minimum P.E. ∴ (D). |