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InterviewSolution
| 1. |
A charge particle q is projected in an electric field produced by a fixed point charge Q as shown in figure. Mark the correct statements. |
| Answer» <html><body><p>The path taken by q is a straight line. <br/>Thepath taken by q is not a straight line. <br/>The <a href="https://interviewquestions.tuteehub.com/tag/minimum-561095" style="font-weight:bold;" target="_blank" title="Click to know more about MINIMUM">MINIMUM</a> distance between the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> particles is `((<a href="https://interviewquestions.tuteehub.com/tag/qq-610902" style="font-weight:bold;" target="_blank" title="Click to know more about QQ">QQ</a>)/(2piepsilon_0) + sqrt((qQ)/(2piepsilon_0) + 4m^2u^2a^2))/(2m u^2)`<br/>Velocity of the particle q is changing in both magnitude and direction. </p>Solution :b.,c.,d. <br/> The path traced by `q` is <a href="https://interviewquestions.tuteehub.com/tag/shown-1206565" style="font-weight:bold;" target="_blank" title="Click to know more about SHOWN">SHOWN</a> in Fig. `SAI.58`, The path is curvilinear and acceleration is due to the force exerted by `Q` on `q`. The separation between them is minimum if relative velocity of the particle along the line joining them is zero. Let `d` be the minimum separation between them. As torque about `Q` is zero, angular momentum remains conserved. <br/> `mua=mvd` or `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>=(ua)/(d)` <br/> From energy conservation law, `(mu^(2))/(2)=(mv^(2))/(2)+(1)/(4piepsilon_(0))(qQ)/(d)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V03_CA1_E01_116_S01.png" width="80%"/></body></html> | |