Question

A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine the object's acceleration & charge magnitude. Assume the acceleration is due to the E-field (i.e., ignore all other forces).

Answer 1

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

q = 0.0000249 C

or

$q=2.49\times 10^{-5}\ C$

Hence, this is the required solution.