Search any question & find its solution
Question:
Answered & Verified by Expert
Two fixed charges A and B of $5 \mu \mathrm{C}$ each are separated by a distance of $6 \mathrm{~m}$. $\mathrm{C}$ is the mid point of the line joining $A$ and $B$. A charge $Q$ of $-5 \mu \mathrm{C}$ is shot perpendicular to the line joining A and $B$ through $C$ with a kinetic energy of $0.06 \mathrm{~J}$. The charge $\mathrm{Q}$ comes to rest at a point D. The distance $C D$ is
Options:
Solution:
1011 Upvotes
Verified Answer
The correct answer is:
$4 \mathrm{~m}$
By conservation of energy
Loss of PE = Gain in KE
$\frac{2 \mathrm{a}^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\sqrt{\mathrm{r}^{2}+\mathrm{x}^{2}}}\right)=\mathrm{K}$
or $\begin{aligned} 2 \times 9 \times 10^{9} \times &\left(5 \times 10^{-6}\right)^{2}\left(\frac{1}{3}-\frac{1}{\sqrt{3^{2}+x^{2}}}\right) \\ &=0.06 \end{aligned}$
or $\quad \frac{1}{3}-\frac{1}{\sqrt{9+x^{2}}}=\frac{2}{15}$
or
$$
x=4 m
$$
Loss of PE = Gain in KE
$\frac{2 \mathrm{a}^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\sqrt{\mathrm{r}^{2}+\mathrm{x}^{2}}}\right)=\mathrm{K}$
or $\begin{aligned} 2 \times 9 \times 10^{9} \times &\left(5 \times 10^{-6}\right)^{2}\left(\frac{1}{3}-\frac{1}{\sqrt{3^{2}+x^{2}}}\right) \\ &=0.06 \end{aligned}$
or $\quad \frac{1}{3}-\frac{1}{\sqrt{9+x^{2}}}=\frac{2}{15}$
or
$$
x=4 m
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.