Question

What is strength of the electric field between two parallel conducting plates separated by 1.00 cm and having an electric potential difference between them of 15,000 V?

Answer 1

Answer:

Electric field will be $1.5\times 10^{6}V/m$

Explanation:

We have given potential difference between the two plates = 15000 volt

Distance between the plates = $d=1cm=10^{-2}m$

We have to find the electric field strength between the plates

Potential difference between the plates is equal to $V=Ed$

So electric field strength will be $E=\frac{V}{d}=\frac{15000}{10^{-2}}=1.5\times 10^{6}V/m$

So electric field will be $1.5\times 10^{6}V/m$