Test Index

JAMB Physics 2017 Paper

© africaexams.com

Question : 3

Total: 49

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 $^{-1}$

Solution:

Concept:

The electric field between two parallel plates is given by the ratio of the potential difference to the separation distance.

Explanation:

Convert distance to metres:

35\,\text{cm} = 0.35\,\text{m}

.

Use the formula:

E = \frac{V}{d}

.

Substitute the given values:

E = \frac{6.5}{0.35}

.

Calculate the electric field intensity:

E = 18.57\,\text{NC}^{-1}

.

Answer:

Option A:

18.57\,\text{NC}^{-1}

.

© africaexams.com

Go to Question:

Prev Question

Next Question