Question

two parallel plates have potential difference of 10v between them if the blades are 0.5 mm apart what will be the strength of electric charge?​

Answer 1

Explanation:

electric cell having potential difference 2v each have been connected in the form of battery what will be the total potential difference of the battery is both cases

Answer 2

Answer:

$2\times 10^4 V/m$

Explanation:

We are given that potential difference between two parallel plates =10 V

Distance between blades=0.5 mm=$0.5\times 10^{-3}m$

We have to find the strength of electric charge .

We know that

Electric field ,E=$\frac{V}{d}$

Where V=Potential difference

d=Distance between two plates

Substitute the values in the given formula

Electric field,E=$\frac{10}{0.5\times 10^{-3}}=\frac{10^2\times 10^3}{5}=2\times 10^4 V/m$

Hence, the electric field strength of electric charge=$2\times 10^4 V/m$