16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4ab,{"@context":"https://schema.org","@type":"Quiz","name":"Aiims 2018 Electrostatics PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/aiims/questions/2018/physics/electrostatics/1lrnkillk","educationalLevel":"Entrance exam preparation","about":"Aiims / Physics / Electrostatics","hasPart":{"@type":"Question","name":"Aiims 2018 Electrostatics PYQ with Solution","text":"A half ring of radius R has a charge of \\lambda per unit length. The electric force on 1 \\mathrm{C} charged placed at the centre is","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"C"},"comment":{"@type":"Comment","text":"As, R be the radius of the ring. Consider a small strip of length d l having charge d q lying at an angle \\theta. d l=R d \\theta Charge on d l=\\lambda R d \\theta Force at 1 \\mathrm{~C} due to \\mathrm{dl} =\\frac{k \\lambda R d \\theta}{R^2}=\\frac{k \\lambda}{R} d \\theta=d F We need to consider only the component d F \\cos \\theta, as the component d F \\sin \\theta will cancel out because of the symmetrical element d l. The total force on 1 \\mathrm{C} is \\begin{aligned} F & =\\int_\\limits{-\\pi / 2}^{\\pi..."}}}18:T4d2,

As, $$R$$ be the radius of the ring. Consider a small strip of length $$d l$$ having charge $$d q$$ lying at an angle $$\theta$$.

$$d l=R d \theta$$

Charge on $$d l=\lambda R d \theta$$

Force at $$1 \mathrm{~C}$$ due to $$\mathrm{dl}$$

$$=\frac{k \lambda R d \theta}{R^2}=\frac{k \lambda}{R} d \theta=d F$$

Electrostatics

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