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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 9 April · Morning Shift · Physics · Electrostatics

At the centre of a half ring of radius R=10 ~cm and linear charge...

At the centre of a half ring of radius R=10 cm\mathrm{R}=10 \mathrm{~cm} and linear charge density 4nCm−14 \mathrm{n} \mathrm{C} \mathrm{m}^{-1}, the potential is xπVx \pi \mathrm{V}. The value of xx is _______

Correct answer

0

Step-by-step solution

Potential at centre of half ring V=KQRV=KλπRRV=Kλπ⇒V=9×109×4×10−9πV=36π\begin{aligned} & V=\frac{K Q}{R} \\ & V=\frac{K \lambda \pi R}{R} \\ & V=K \lambda \pi \Rightarrow V=9 \times 10^9 \times 4 \times 10^{-9} \pi \\ & V=36 \pi\end{aligned}
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