Question

In an insulating medium (K = 1) volumetric charge density varies with y-coordinates according to the law r = a.y. A particle of mass m having positive charge q is at point A(0, y₀) and projected with velocity $\overrightarrow{v} = v_{0}\widehat{i}$as shown in figure. At y = 0 electric field is zero. Neglect the gravity and fractional resistance, the slope of trajectory of the particle as a function of y (E is only along y-axis) is –

• A. $\sqrt{\frac{qa}{m\varepsilon_{0}v_{0}^{2}}(y^{3}–y_{0}^{3})}$
• B. $\sqrt{\frac{qa}{3m\varepsilon_{0}v_{0}^{2}}(y^{3}–y_{0}^{3})}$
• C. $\sqrt{\frac{qa(y^{3}–y_{0}^{3})}{5m\varepsilon_{0}v_{0}^{2}}}$
• D. $\sqrt{\frac{qa(y^{3}–y_{0}^{3})}{2m\varepsilon_{0}v_{0}^{2}}}$

Answer 1

Answer: (B)

$\sqrt{\frac{qa}{3m\varepsilon_{0}v_{0}^{2}}(y^{3}–y_{0}^{3})}$

Answer 2

Gauss law,

(E + dE) A – EA = $\frac{\rho Ady}{\varepsilon_{0}}$

dE = $\frac{\rho dy}{\varepsilon_{0}}$= $\frac{aydy}{\varepsilon_{0}}$

$\int_{0}^{E}{dEy}$= $\int_{0}^{y}{ydy}$

E_y = $\frac{ay^{2}}{2\varepsilon_{0}}$

E is not present along x-axis

a = $\frac{qE}{m}$= $\frac{qay^{2}}{2m\varepsilon_{0}}$

$v\frac{dv}{dy}$ = $\frac{qay^{2}}{2m\varepsilon_{0}}$

Integrating

$\int_{}^{}{vdv = \frac{qa}{2m\varepsilon_{0}}\int_{y_{0}}^{y}{y^{2}dy}}$v = calculated

Slope = $\frac{v_{y}}{v_{x}}$ = $\frac{v}{v_{0}}$