Question

Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is
A) 800 N/C
B) 1600 N/C
C) 2400 N/C
D) 5000 N/C
E) 20000 N/C

Answer 1

Answer:

E)  E₂= 20000 N/C

Explanation:

The electric field between two parallel conductive plates is thus calculated

E= V/d

Where:

E: Electric field (N/C)

V:  voltage (V)

d : distance between the plates (m)

Problem development

E₁= V₁/d₁  = 2000 N/C

E₂= V₂/d₂

$E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }$

$E_{2} = 10(\frac{V_{1} }{d_{1} })$

E₂= 10* (2000)

E₂= 20000 N/C