Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
-
18.57NC\(^{-1}\)
- 53.06N C\(^{-1}\)
- 2.28NC\(^{-1}\)
- 0.80NC\(^{-1}\)
The correct answer is: A
Explanation
Electric Field Intensity (E) = \(\frac{v}{d}\)= \(\frac{\text{potential difference}}{\text{distance}}\)
= \(\frac{6.5v}{35cm}\)
= \(\frac{6.5}{35 \times 10^{- 2}}\)
= 18.57NC\(^{-1}\)
There is an explanation video available .