Question

An electron (mass$=9.1\times {10}^{-31}\mathrm{kg}$) is sent into an electric field of intensity $9.1\times {10}^6\mathrm{N}/\mathrm{C}$. The acceleration produced is

• A. $1.6\times {10}^{18}{\mathrm{ms}}^{-2}$
• B. $1.6\times {10}^6{\mathrm{ms}}^{-2}$
• C. $1.6\times {10}^{-18}{\mathrm{ms}}^{-2}$
• D. $1.6\times {10}^{-6}{\mathrm{ms}}^{-2}$

Answer 1

The correct option is A

$1.6\times {10}^{18}{\mathrm{ms}}^{-2}$

Step 1: Given data

Mass of electron, $m=9.1\times {10}^{-31}\mathrm{kg}$

Intensity of electric field, $q_e=9.1\times {10}^6\mathrm{N}/\mathrm{C}$

Step 2: Formulas used

From Newton's second law of motion, we know that

$\mathrm{F}=ma$

Where, $\mathrm{F}$ is the force on the object, $m$ is the mass of the object and $a$ is the acceleration produced in the object.

We know that, the force on a charge exerted by an electric field is given as,

$\mathrm{F}=q_{\mathit{e}}E$

Where $q_e$ is the intensity of the field and $E$ is the charge of the body.

Step 3: Calculate acceleration on the charge

By modifying first equation,

$a=\frac{\mathrm{F}}{m}$

Substituting the value of $\mathrm{F}$ in the above equation with the expression from the second equation, we get

$\begin{array}{cc}& a=\frac{q_{e}E}{m}\end{array}$
$\begin{array}{cc}\Rightarrow & a=\frac{\left(9.1\times {10}^6\right)\times \left(1.6\times {10}^{-19}\right)}{9.1\times {10}^{-31}}\end{array}$ $[\because$charge of an electron$=1.6\times {10}^{-16}\mathrm{C}]$
$\begin{array}{cc}\therefore & a=1.6\times {10}^{18}{\mathrm{ms}}^{-2}\end{array}$

Therefore, the acceleration gained by the electron is $1.6\times {10}^{18}{\mathrm{ms}}^{-2}$.

Hence, option A is correct.