Question

Two large conducting plates are placed parallel to each other with a separation of 2⋅00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2⋅00 microseconds. Find the surface charge density on the inner surfaces.

Answer 1

Answer:

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Answer 2

The surface charge density on the inner surfaces is 0.503$\times$10−12 C/m2.

Explanation:

• The distance travelled by the electron, d= 2 cm
• The needed time taken to cross the region, t = 2$\times$10−-6 s
• Say, the surface charge density at the conducting plates be σ.
• Say,  the acceleration of the electron be a.

When we apply the 2nd equation of motion, we get:

d=12 at2         ⇒a=2dt2           d=12 at2         ⇒a=2dt2            

The Coulombic force provides this acceleration. So,

 a=qE/m=2d/t2

⇒E=2 md/qt2  

E=2$\times$(9.1$\times$10−31)$\times$(2$\times$10−2)/(1.6$\times$10−19)$\times$(4$\times$10−12)

E=5.6875$\times$10−2 N/C  

We know that electric field due to a plate,

E=σ/∈0

⇒σ = ∈0E

⇒σ=(8.85$\times$10−12)$\times$(5.68$\times$10−2) C/m2

⇒σ=50.33$\times$10−14 C/m2=0.503$\times$10−12 C/m2