Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
A
18.57NC\(^{-1}\)
B
53.06N C\(^{-1}\)
C
2.28NC\(^{-1}\)
D
0.80NC\(^{-1}\)
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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\(^{-1}\)
Source: Myschool Past Questions Archive (Record #46376)