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WAEC Physics 2016 Paper

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Question : 35

Total: 47

The p.d across a parallel-plate capacitor is 10³V. If the distance between the two plates is 10cm. Calculate the magnitude of the electric field strength between the plates.

1.0 x 10⁴Vm^-1
1.0 x 10⁵Vm^-1
1.5 x 10⁵Vm^-1
2.0 x 10⁵Vm^-1

Solution:

Concept:

The electric field strength between parallel plates is the potential difference per unit separation.

Explanation:

Given potential difference,

V = 10^3 \, \text{V}

.

Convert distance to metres:

d = 10 \, \text{cm} = 0.1 \, \text{m}

.

Electric field strength,

E = \frac{V}{d}

.

Substitute the values:

E = \frac{10^3}{0.1}

.

So,

E = 10000 \, \text{Vm}^{-1} = 1.0 \times 10^4 \, \text{Vm}^{-1}

.

Answer:

E = 1.0 \times 10^4 \, \text{Vm}^{-1}

.

Correct option: A.

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