Question

A uniform electric field of magnitude $250\text{V/m}$ is directed in the positive 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 by us in moving the charge particle to new position

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

Answer 1

The correct option is A

$-0.6m\text{J}$
Given that,
Electric field, $\overrightarrow{E}=250\text{V/m}$

Charge, $q=12\mu \text{C}$

If we move along $y-$ axis then there is no change in potential because whole plane is at same potential but as we move along x-axis then there will be change in potential.

Let the potential at the origin be $V_1$ and at point ($20.0\text{cm},5.0\text{cm}$) is $V_2$
$V_2-V_1=\Delta V=-\overrightarrow{E}.\overrightarrow{d}$
$\Rightarrow \Delta V=-250\times 20\times {10}^{-2}$
$\Rightarrow \Delta V=-50\text{V}$

Work done to move the charge $q$ be $W=q\Delta V$

Substituting the given data we get,

$W=12\times {10}^{-6}\times -50=-0.6m\text{J}$

Hence, option (a) is the correct answer.