Question

2 parallel plates 5 cm apart and a potential difference of 60v is set up across them find the electric field intensity between the 2 plates​

Answer 1

Relation between potential difference and electric field:

$E =-\frac{ \delta V }{\delta r}$

E = -V/r

E= -60/0.05

E = -1200 N/C

Answer 2

Answer:

Explanation:

Distance r= 5 cm

Potential difference= 60 volt

E= electric field intensity

E = -dv/ dr

-60/0.5(in meters)

= -1200V/m