Question

two parallel plates are 0.03 M apart the electric field intensity between them is 3000 Newton per coulomb calculate the potential difference between the plates​

Answer 1

two parallel plates are 0.03 M apart the electric field intensity between them is 3000 Newton per coulomb.

we have to potential difference between the plates

given, distance between two parallel plates , d = 0.03 m

electric field intensity between plates , E = 3000 N/C

we know, potential difference = electric field intensity within two parallel plates × distance between two parallel plates

or, V = E × d = 3000 × 0.03

= 30 × 3 = 90V

hence, potential difference between the plates , V = 90V

Answer 2

Answer:90 v

Explanation:

two parallel plates are 0.03 M apart the electric field intensity between them is 3000 Newton per coulomb.

we have to potential difference between the plates

given, distance between two parallel plates , d = 0.03 m

electric field intensity between plates , E = 3000 N/C

we know, potential difference = electric field intensity within two parallel plates × distance between two parallel plates

or, V = E × d = 3000 × 0.03

= 30 × 3 = 90V

hence, potential difference between the plates , V = 90V