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Below are the choices that can be found from other sources:
A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2
B. +2.7 × 10−9 C/m2; −2.7 × 10−9 C/m2
C. −5.3 × 10−9 C/m2; +5.3 × 10−9 C/m2
D. +5.3 × 10−9 C/m2; −5.3 × 10−9 C/m2
E. 0; 0
The answer is A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2
Below are the choices that can be found from other sources:
A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2
B. +2.7 × 10−9 C/m2; −2.7 × 10−9 C/m2
C. −5.3 × 10−9 C/m2; +5.3 × 10−9 C/m2
D. +5.3 × 10−9 C/m2; −5.3 × 10−9 C/m2
E. 0; 0
The answer is A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2
Answer:
[tex]\sigma_{left} = -2.7 \times 10^{-10} C/m^2[/tex]
[tex]\sigma_{right} = 2.7 \times 10^{-10} C/m^2[/tex]
Explanation:
As we know that electric field due to a large sheet is given by the formula
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
here we know that
[tex]\sigma [/tex] = charge density
now we know that here the field is given for a point between two charged plates
so net electric field is given as
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
[tex]\sigma = E \epsilon_0[/tex]
[tex]\sigma = (30)(8.85 \times 10^{-12})[/tex]
[tex]\sigma = 2.7 \times 10^{-10} C/m^2[/tex]
now here electric field is indicating towards left so here charge density on left side is negative while on right side it must be positive
