Question

In a region where there is a uniform electric field that is upward and has a magnitude of 3.60 x 10⁴ N/C, a small object is projected upward with an initial speed of 1.92 m/s. The object travels upward a distance of 6.98 cm in 0.200 s. What is the object’s charge-to-mass ratio q/m? Assume g = 9.80 m/s ², and ignore air resistance.

Short answer

The charge to mass ratio($\frac{q}{m}$) is equal to $-1.64\times {10}^{-4}\mathrm{C}/\mathrm{Kg}$

Step-by-step solution

About Electric field

We are given the value of the uniform electric field $\mathrm{E}=3.60\times {10}^4\mathrm{N}/\mathrm{C}$that is upward. Als, we are given the initial speed of the object ${\mathrm{v}}_0=1.92\mathrm{m}/\mathrm{s}$and travels upward a distance d=6.98cm = 0.0698m in time t=0.200 second

To express a relationship between the mass and charge of the object, we will use Newton's law which is given as

F=ma

And as the electric field exerted a force on the object, this will made the force that is exerted on the object is given by

$$" width="9">F=qE=maE (1)

localid="1664523323203" $$\begin{gathered} \mathrm{qE}={\mathrm{ma}}_{\mathrm{E}} \\ \frac{\mathrm{q}}{\mathrm{m}}=\frac{{\mathrm{a}}_{\mathrm{E}}}{\mathrm{E}} \end{gathered}$$

Where m is the mass of the object, is the acceleration due to the force of the electric field. Now our target is to find the value to complete the calculation. The object travels upward with an initial speed, so we can use Newton's law of motion to find the acceleration of the object.

$\mathrm{d}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (2)

$a=2\frac{(d-v_{0}t)}{t^2}$

Now let us put our value for d,$d_1{v}_0$and t into equation (2) to find a

$a=2\frac{(d-v_{0}t)}{t^2}$

$$\begin{gathered} 2\frac{\left(0.0698m-1.92m/s\times 0.200s\right)}{{\left(0.200s\right)}^2} \\ =-15.71m/s^2 \end{gathered}$$

Calculations

Another acceleration that affects the motion of the object, is the gravitational acceleration g, so(a) represents the total acceleration of the object and it is given by

$$\begin{gathered} \mathrm{a}=\mathrm{g}+{\mathrm{a}}_{\mathrm{E}} \\ {\mathrm{a}}_{\mathrm{E}}=\mathrm{a}-\mathrm{g} \end{gathered}$$

Where g is equal to $9.8\mathrm{m}/{\mathrm{s}}^2$as the object is downward. Now we can put $a_E$by

${\mathrm{a}}_{\mathrm{E}}=\mathrm{a}-\mathrm{g}=-15.71\mathrm{m}/{\mathrm{s}}^2-(-9.8\mathrm{m}/{\mathrm{s}}^2)=-5.91\mathrm{m}/{\mathrm{s}}^2$

Now we can put our value for $a_E$and E into equation (1) to get the ratio q/m

$\frac{\mathrm{q}}{\mathrm{m}}=\frac{{\mathrm{a}}_{\mathrm{E}}}{\mathrm{E}}=\frac{-5.91\mathrm{m}/{\mathrm{s}}^2}{3.60\times {10}^4\mathrm{N}/\mathrm{C}}=-1.64\times {10}^{-4}\mathrm{C}/\mathrm{Kg}$