The electric potential varies in space accord | vedclass

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(D) The electric field components are given by $E_x = -\frac{\partial V}{\partial x} = -3 \, V/m$ and $E_y = -\frac{\partial V}{\partial y} = -4 \, V/

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) The electric field components are given by $E_x = -\frac{\partial V}{\partial x} = -3 \, V/m$ and $E_y = -\frac{\partial V}{\partial y} = -4 \, V/m$.The acceleration components are $a_x = \frac{q E_x}{m} = \frac{1 	imes 10^{-6} 	imes (-3)}{0.1} = -3 	imes 10^{-5} \, m/s^2$ and $a_y = \frac{q E_y}{m} = \frac{1 	imes 10^{-6} 	imes (-4)}{0.1} = -4 	imes 10^{-5} \, m/s^2$.The particle crosses the $x$-axis when its $y$-coordinate becomes $0$. The initial $y$-coordinate is $y_0 = 3.2 \, m$ and initial velocity $u_y = 0$.Using the equation of motion $y = y_0 + u_y t + \frac{1}{2} a_y t^2$,we set $y = 0$:$0 = 3.2 + 0 - \frac{1}{2} 	imes (4 	imes 10^{-5}) 	imes t^2$.$3.2 = 2 	imes 10^{-5} 	imes t^2$.$t^2 = \frac{3.2}{2 	imes 10^{-5}} = 1.6 	imes 10^5 = 16 	imes 10^4$.$t = \sqrt{16 	imes 10^4} = 400 \, s$.

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