Question

A uniform electric field of magnitude $250\text{V/m}$ is directed in the positive $\text{x-}$direction. A $+12\mu \text{C}$ charge moves from the origin to the point $(x,y)=(20.0\text{cm},$$5.0\text{cm})$. What is the total work done required in moving the charge particle to the new position?

• A. $-0.6\text{mJ}$
• B. $0.4\text{mJ}$
• C. $4\mu \text{J}$
• D. $6\mu \text{J}$

Answer 1

The correct option is A $-0.6\text{mJ}$

Given,

Electric field, $E=250\text{V/m}$

Charge, $q=12\mu \text{C}=12\times {10}^{-6}\text{C}$,

If we move along $\text{y}-$axis then there is no change in potential because whole plane is at same potential but as we move along $\text{x}-$axis then there will be change in potential so, change in potential energy will be

$V_2-V_1=\overrightarrow{E.}\overrightarrow{d}$

Substituting the values given in the question,

$\Rightarrow V_2-V_1=-250\times 20\times {10}^{-2}$

$\Rightarrow V_2-V_1=-50\text{V}$

So, workdone in moving charge $q$, from origin to new position,

$W=q(V_2-V_1)$

$\Rightarrow W=12\times {10}^{-6}\times (-50)$

$\Rightarrow W=-6\times {10}^{-4}\text{J}$

$\Rightarrow W=-0.6\text{mJ}$

Hence, option (a) is correct asnwer.