Question

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

• A. $1.6\times {10}^{18}m/s^2$
• B. $1.6\times {10}^{6}m/s^2$
• C. $1.6\times {10}^{-18}m/s^2$
• D. $1.6\times {10}^{-6}m/s^2$

Answer 1

The correct option is A $1.6\times {10}^{18}m/s^2$

$\text{Step 1 : Force on electron due to electric field}$

$F=q_{e}E$ $....\left(1\right)$

$\text{Step 2 : Acceleration produced:}$

We know that, $F=ma$

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

Using equation $\left(1\right)$

$a=\frac{q_{e}E}{m_e}$

$=\frac{(9.1\times {10}^{6}N/C)\times (1.6\times {10}^{-19}C)}{9.1\times {10}^{-31}Kg}$ $=1.6\times {10}^{18}m/s^2$

Hence, acceleration of electron is $1.6\times {10}^{18}m/s^2$