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})
Correct Answer: Option 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 below.