Question

The electric potential varies in space according to the relation V = 3x + 4 y. A particle of mass 10 kg starts from rest from point (2, 3.2) m under the influence of this field. The velocity of the particle when it crosses the x-axis is equal t o $x\times {10}^{-3}m/s$. The charge on the particle is +1$\mu$C Assume V (x, y) are in SI units. Find the value of $x$.

Answer 1

V=3x+4y

Initial position is $(2,3.2)$, hence,$V_1=3\times 2+4\times 3.2=18.8V$

Now, $E_x=\frac{-\partial V}{\partial x}=-3V$

and $E_y=\frac{-\partial V}{\partial y}=-4V$

Hence, $\overrightarrow{E}=-3\hat{i}-4\hat{i}$

Hence, path of charge will be along this vector as shown in figure.

Now, equation of line of travel of charge:-

slope$=\frac{4}{3}$

$y-3.2=\frac{4}{3}(x-2)$

When, point passes through X-axis, $y=0$:-

$⟹-3.2=\frac{4}{3}(x-2)$

$⟹x=-0.4$

Here, potential=$V_2=3\times (-0.4)+4\times 0=-1.2V$

Now, gain in kinetic energy= loss in potential energy

$⟹K_2-K_1=V_1-V_2$

$⟹\frac{1}{2}m{v}^2=q\{18.8-(-1.2\left)\right\}=20q$

$⟹v^2=\frac{40\times {10}^{-6}}{10}$

$⟹v^2=4\times {10}^{-6}$

$⟹v=2\times {10}^{-3}m/s$