Physics · Electrostatics
JEE Main 2024 — 6 April, Shift 1 — Question 51
Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is . The value of x is
(all quantities are measured in SI units).
Answer: 2
Numerical answer — enter this value.
Step-by-step solution
\begin{array}{*{35}{r}}{{{\text{\overrightarrow E}}}_{\text{p}}} & ~=\left( \frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{2\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}} \right)\left( -\overset{\text{}}{\mathop{\text{i}}}\, \right) \\{} & ~=-\frac{2\sigma }{{{\varepsilon }_{0}}}\overset{\text{}}{\mathop{\text{i}}}\, \\\end{array}
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 6 April, Shift 1
- Subject
- Physics
- Chapter
- Electrostatics
- Topic
- Electrostatic Field & motion of charge