TS EAMCET · Physics · Current Electricity

A semi-infinite non conducting rod lies along $+ x$ -axis with its left end at the origin. The rod has uniform linear charge density $λ .$ The magnitude of electric field $| \vec{E} |$ at a point on the $y$ -axis at the distance $L$ from the origin, will be

A $\frac{λ}{4 π ε_{0} L}$
B $\frac{λ}{2 π ε_{0} L}$
C $\frac{λ}{2 \sqrt{2} π ε_{0} L}$
D $\frac{\sqrt{2} λ}{π ε_{0} L}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{λ}{2 \sqrt{2} π ε_{0} L}$

Step-by-step Solution

Detailed explanation

Consider an infinitesimal section of the rod of length d x , a distance x from the left end, as shown in the figure below. It contains charge d q = λ d x and is a distance r from L .Magnitude of the fieldproduces by d x at L is given by d E = k λ d x r 2 .The x and…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A semi-infinite non conducting rod lies along +x-axis with its left end at the origin. The rod has uniform linear charge density λ. The magnitude of electric field |E→| at a point on the y-axis at the distance L from the origin, will be

Step-by-step Solution

A semi-infinite non conducting rod lies along $+ x$ -axis with its left end at the origin. The rod has uniform linear charge density $λ .$ The magnitude of electric field $| \vec{E} |$ at a point on the $y$ -axis at the distance $L$ from the origin, will be