Chapter 21, Problem 103A

To determine

The vertical force on the ink drops, vertical acceleration of the ink drops, the time up to which an ink drop will be there in between the plates and the vertical displacement of the ink drops when they leave the plates.

Answer to Problem 103A

The vertical force on the ink drops is $1.2\times {\text{10}}^{-10}\text{ N}$ .

The vertical acceleration of the ink drops is $1.2\times {\text{10}}^3{\text{ m/s}}^2$ .

The time up to which an ink drop will be there in between the plates is $\text{1}\text{.0}\times {\text{10}}^{-3}\text{ s}$ .

The vertical displacement of the ink drops when they leave the plates $6.0\times {\text{10}}^{-4}\text{ m}$

Explanation of Solution

Given:

Electric field strength is $E=1.2\times {\text{10}}^6\text{ N/C}$

Charge in the capacitor is $q=1.0\times {\text{10}}^{-16}\text{ C}$

Figure 1

Mass of an ink drop is $\text{m = 0}\text{.1 ng = 1}\times {\text{10}}^{-13}\text{ kg}$

The length of the plate is $L=1.5\text{ cm = 1}\text{.5}\times {\text{10}}^{-2}\text{ m}$

Velocity of an ink drop between the plates is $v\text{ = 15 m/s}$

Formula used:

The electric field strength $\left(E\right)$ is given by,

  $E=\frac{F}{q}$

  $\Rightarrow F=Eq$ $\left(1\right)$

Where, $F$ is the electric force

  $q$ is the electric charge

Force equation (Newton’s second law of motion) is,

  $F=ma$

  $\Rightarrow a=\frac{F}{m}$ $\left(2\right)$

Where, $a$ is acceleration of an ink drop

  $m$ is mass of an ink drop

The time up to which an ink drop will be there in between the plates can be calculated using the relation,

  $t\text{ = }\frac{L}{v}$ $\left(3\right)$

Where, $L$ is the length of the plates

  $v$ is the velocity of an ink drop

The vertical displacement of the ink drops when they leave the plates can be calculated using the equation,

  $s=ut+\frac{1}{2}a{t}^2$ $\left(4\right)$

Where, $u$ is the initial velocity of the ink drop, $u=0.0\text{ m/s}$

Where, $a$ is acceleration of an ink drop

  $t$ time taken by an ink drop to pass through the plates.

Calculation:

To find the force, substitute the numerical values in equation $\left(1\right)$ ,

  $F=\left(1.2\times {\text{10}}^6\text{ N/C}\right)\left(1.0\times {\text{10}}^{-16}\text{ C}\right)$

  $=1.2\times {\text{10}}^{-10}\text{ N}$

$F$ is force on an ink drop, $F=1.2\times {\text{10}}^{-10}\text{ N}$

To find the acceleration, substitute the numerical values in equation $\left(2\right)$ ,

  $a=\frac{1.2\times {\text{10}}^{-10}\text{ N}}{\text{1}\times {\text{10}}^{-13}\text{ kg}}=\frac{1.2\times {\text{10}}^{-10}{\text{ kgm/s}}^2}{\text{1}\times {\text{10}}^{-13}\text{ kg}}=1.2\times {\text{10}}^3{\text{ m/s}}^2$

To find the time, substitute the numerical values in equation $\left(3\right)$ ,

  $t=\frac{\text{ 1}\text{.5}\times {\text{10}}^{-2}\text{ m}}{\text{15 m/s}}=\text{1}\text{.0}\times {\text{10}}^{-3}\text{ s}$

To find the distance, substitute the numerical values in equation $\left(4\right)$ ,

  $s=\left(0.0\text{ m/s}\right)\left(\text{1}\text{.0}\times {\text{10}}^{-3}\text{ s}\right)+\frac{1}{2}\left(1.2\times {\text{10}}^3{\text{ m/s}}^2\right){\left(\text{1}\text{.0}\times {\text{10}}^{-3}\text{ s}\right)}^2$

  $=\frac{1}{2}\left(1.2\times {\text{10}}^3{\text{ m/s}}^2\right)\left(\text{1}\text{.0}\times {\text{10}}^{-6}{\text{ s}}^2\right)$

  $=6.0\times {\text{10}}^{-4}\text{ m}$

Conclusion:

The vertical force on the ink drops is $1.2\times {\text{10}}^{-10}\text{ N}$ .

The vertical acceleration of the ink drops is $1.2\times {\text{10}}^3{\text{ m/s}}^2$ .

The time up to which an ink drop will be there in between the plates is $\text{1}\text{.0}\times {\text{10}}^{-3}\text{ s}$ .

The vertical displacement of the ink drops when they leave the plates $6.0\times {\text{10}}^{-4}\text{ m}$