KCET · Physics · Electrostatics
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
- A \(4 \mathrm{~m}\)
- B \(3 \mathrm{~m}\)
- C \(\sqrt{3} \mathrm{~m}\)
- D \(3 \sqrt{3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(4 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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
\(2 \times 9 \times 10^{9} \times \left(5 \times 10^{-6}\right)^{2}\left(\frac{1}{3}-\frac{1}{\sqrt{3^{2}+x^{2}}}\right) \)
\( =0.06\)
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
\(2 \times 9 \times 10^{9} \times \left(5 \times 10^{-6}\right)^{2}\left(\frac{1}{3}-\frac{1}{\sqrt{3^{2}+x^{2}}}\right) \)
\( =0.06\)
or \(\quad \frac{1}{3}-\frac{1}{\sqrt{9+x^{2}}}=\frac{2}{15}\)
or
\(x=4 m\)
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