Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
Answer Details
The electric field intensity E between two parallel plates of potential difference V separated by a distance d is given by E = V/d.
Substituting the given values, we have E = 6.5V/0.35m = 18.57 NC^−1.
Therefore, the electric field intensity between the two plates is 18.57 NC^−1.
Answer: 18.57NC^−1