Let be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions , and is [2023]

Question

Let $σ$ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions $E_{I}$ , $E_{I I}$ and $E_{I I I}$ is [2023]

Smart Achievers · Accepted Answer

(4)

$Assuming RHS to be \hat{n}$

${\vec{E}}_{I} = \frac{σ}{2 ε_{0}} (- \hat{n}) + \frac{σ}{2 ε_{0}} (- \hat{n}) = - \frac{σ}{ε_{0}} \hat{n}$

${\vec{E}}_{I I} = 0$

${\vec{E}}_{I I I} = \frac{σ}{2 ε_{0}} (\hat{n}) + \frac{σ}{2 ε_{0}} (\hat{n}) = \frac{σ}{ε_{0}} (\hat{n})$

Solution

Q.
Let $σ$ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions $E_{I}$ , $E_{I I}$ and $E_{I I I}$ is [2023]

1 ${\vec{E}}_{I} = \frac{2 σ}{ε_{0}} \hat{n}, {\vec{E}}_{I I} = 0, {\vec{E}}_{I I I} = \frac{2 σ}{ε_{0}} \hat{n}$

2 ${\vec{E}}_{I} = 0, {\vec{E}}_{I I} = \frac{σ}{ε_{0}} \hat{n}, {\vec{E}}_{I I I} = 0$

3 ${\vec{E}}_{I} = \frac{σ}{2 ε_{0}} \hat{n}, {\vec{E}}_{I I} = 0, {\vec{E}}_{I I I} = \frac{σ}{2 ε_{0}} \hat{n}$

4 ${\vec{E}}_{I} = - \frac{σ}{ε_{0}} \hat{n}, {\vec{E}}_{I I} = 0, {\vec{E}}_{I I I} = \frac{σ}{ε_{0}} \hat{n}$

Ans.
(4)

$Assuming RHS to be \hat{n}$

${\vec{E}}_{I} = \frac{σ}{2 ε_{0}} (- \hat{n}) + \frac{σ}{2 ε_{0}} (- \hat{n}) = - \frac{σ}{ε_{0}} \hat{n}$

${\vec{E}}_{I I} = 0$

${\vec{E}}_{I I I} = \frac{σ}{2 ε_{0}} (\hat{n}) + \frac{σ}{2 ε_{0}} (\hat{n}) = \frac{σ}{ε_{0}} (\hat{n})$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let σ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions EI, EII and EIII is [2023]

1 E→I=2σε0n^, E→II=0, E→III=2σε0n^

2 E→I=0, E→II=σε0n^, E→III=0

3 E→I=σ2ε0n^, E→II=0, E→III=σ2ε0n^

4 E→I=-σε0n^, E→II=0, E→III=σε0n^

Ans. (4) Assuming RHS to be n^ E→I=σ2ε0(-n^)+σ2ε0(-n^)=-σε0n^ E→II=0 E→III=σ2ε0(n^)+σ2ε0(n^)=σε0(n^)