Question

A charge of magnitude '2e' and mass '4m' is moving in an electric field $\vec{E}$. The acceleration imparted to the above charge is

• A. $\frac{2m}{3Ee}$
• B. $\frac{Ee}{2m}$
• C. $\frac{2Ee}{3m}$
• D. $\frac{3m}{2Ee}$

Hint:

The simple base equation for acceleration inside an electric field is $a = \frac{qE}{m}$. Since the charge is doubled ($\times 2$) but the mass is quadrupled ($\times 4$), the base value gets scaled by a factor of $\frac{2}{4} = \frac{1}{2}$. This leads you directly to $\frac{Ee}{2m}$ without writing out lines of scratch work!

Answer 1

The Correct Option is B

The electrostatic force ($F$) experienced by any point charge $q$ placed inside an external electric field $\vec{E}$ is given by: $$F = qE$$ Substituting the given charge magnitude parameter $q = 2e$: $$F = 2eE$$ According to Newton's second law of motion, the acceleration ($a$) experienced by an object is its net force divided by its mass ($m_{\text{mass}}$): $$a = \frac{F}{m_{\text{mass}}}$$ Substituting our force value along with the given particle mass parameter $m_{\text{mass}} = 4m$: $$a = \frac{2eE}{4m} = \frac{eE}{2m} = \frac{Ee}{2m}$$
Final Answer:
The acceleration imparted to the charge is $\frac{Ee}{2m}\boxext$, which corresponds to option (B).