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A $2 \mathrm{~g}$ object, located in a region of uniform electric field $\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}$ carries a charge $Q$. The object released from rest at $x=0$, has a kinetic energy of $0.12 \mathrm{~J}$ at $x=0.5 \mathrm{~m}$. Then, $Q$ is
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The correct answer is:
$800 \mu \mathrm{C}$
Given, $m=2 g=2 \times 10^{-3} \mathrm{~kg}$
$\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}$
At, $x=0,(\mathrm{KE})_{1}=0$ and $x=0.5 \mathrm{~m},(\mathrm{KE})_{2}=0.12 \mathrm{~J}$
From work-energy theorem,
work done $=$ change in kinetic energy
$\begin{aligned}
&\Rightarrow \text { Force } \times \text { displacement }=\Delta \mathrm{KE} \\
&\qquad \begin{aligned}
Q E \times d &=\mathrm{KE}_{2}-\mathrm{KE}_{1} \\
\Rightarrow & Q \times 300 \times 0.5=0.12 \\
\text { or } & Q=\frac{0.12}{300 \times 0.5}=800 \mu \mathrm{C}
\end{aligned}
\end{aligned}$
The charge $Q$ is moving in the direction of electric field. Hence, it is positive.
$\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}$
At, $x=0,(\mathrm{KE})_{1}=0$ and $x=0.5 \mathrm{~m},(\mathrm{KE})_{2}=0.12 \mathrm{~J}$
From work-energy theorem,
work done $=$ change in kinetic energy
$\begin{aligned}
&\Rightarrow \text { Force } \times \text { displacement }=\Delta \mathrm{KE} \\
&\qquad \begin{aligned}
Q E \times d &=\mathrm{KE}_{2}-\mathrm{KE}_{1} \\
\Rightarrow & Q \times 300 \times 0.5=0.12 \\
\text { or } & Q=\frac{0.12}{300 \times 0.5}=800 \mu \mathrm{C}
\end{aligned}
\end{aligned}$
The charge $Q$ is moving in the direction of electric field. Hence, it is positive.
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