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# A mass m = 20 g has a charge q = 3.0 mC. It moves with a velocity of 20 m/s and enters a region of electric field of 80 N/C in the same direction as the velocity of the mass. The velocity of the mass after 3 seconds in this region is:

Last updated date: 16th Jun 2024
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Hint: The acceleration is calculated from the force experienced by the body in the electric field and final velocity is calculated from the equation of motion.
It can be calculated by using Formulas:
1. $a = \dfrac{{qE}}{m}$
2. $v = u + at$

Step by step answer: A body of mass m = 20 g which has a charge q of 3.0 mC i.e., 3 ×10⁻3 C ( since 1 C = 103 mC). The initial velocity (u) is 20 m/s through which it enters a region of electric field. As the body enters the region, it experiences a force due to this electric field (E). The body continues to move in the same direction. So, the final velocity (v) of the body is positive i.e., v is positive.
The force (F) experienced by the mass m is provided by the electric field of the region in which the body enters.
Thus, $F = qE$
$\Rightarrow$ $ma = qE$
[$F = ma$ where a is the acceleration of the body of mass m]
$\Rightarrow$ $a = \dfrac{{qE}}{m}$
$\Rightarrow$ $[ E = 80 N/C and m = 20 × 10⁻3g as 1 g = 10⁻3 kg]$
$\Rightarrow$ a = 12 m/s2
Now, The velocity (v) of the mass after time t = 3 sec is calculated by using the equation
$v = u + at$
$\Rightarrow$ $v = 20 + 12 \times 3$
$\therefore$ $v = 20 + 36 = 56$
Therefore, the final velocity of the mass after 3 sec is 56 m/s.

Note: The particle continues to move in the same direction. So, the sign of final velocity will be positive and hence, the acceleration will be also positive.