A particle of mass 1 , g and charge -0.1 , mu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $m = 1 \, g = 10^{-3} \, kg$, $q = -0.1 \, \mu C = -10^{-7} \, C$, $E = 1 \, kV/cm = 10^5 \, V/m$, $u = 10\sqrt{2} \, m/s$, $	heta = 45^o$

Vedclass · Accepted Answer

Range of the particle is $20 \, m$.

Vedclass · Answer

(B) Given: $m = 1 \, g = 10^{-3} \, kg$, $q = -0.1 \, \mu C = -10^{-7} \, C$, $E = 1 \, kV/cm = 10^5 \, V/m$, $u = 10\sqrt{2} \, m/s$, $	heta = 45^o$, $g = 10 \, m/s^2$.Initial velocity components: $u_x = u \cos 45^o = 10 \, m/s$, $u_y = u \sin 45^o = 10 \, m/s$.Acceleration components: $a_y = -g = -10 \, m/s^2$. Since the charge is negative, the electric force $F_x = qE$ acts in the negative $x$-direction. $a_x = \frac{qE}{m} = \frac{(-10^{-7}) 	imes 10^5}{10^{-3}} = -10 \, m/s^2$.Time of flight $T = \frac{2u_y}{g} = \frac{2 	imes 10}{10} = 2 \, s$. (Statement $A$ is correct).Range $R = u_x T + \frac{1}{2} a_x T^2 = (10)(2) + \frac{1}{2}(-10)(2^2) = 20 - 20 = 0 \, m$. (Statement $B$ is incorrect).Total displacement $S = \sqrt{x^2 + y^2}$. Since $x = R = 0$ and $y = u_y T - \frac{1}{2} g T^2 = 10(2) - 5(4) = 0$, the total displacement is $0 \, m$. (Statement $C$ is correct).Since $a_x = a_y = -10 \, m/s^2$, the ratio $\frac{a_y}{a_x} = 1 = \frac{u_y}{u_x}$, so the particle moves in a straight line. (Statement $D$ is correct).

Similar Questions

$A$ positively charged particle moving along the $x$-axis with a certain velocity enters a uniform electric field directed along the positive $y$-axis. Its

$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.

The magnitude of an electric field which can just suspend a deuteron of mass $3.2 \times 10^{-27} \ kg$ freely in air is

Vedclass Test Series

Exam Paper Generator

Online Exam Module