The potential difference between two parallel plates 1 cm apart is 100V. The electric field strength between them is :

100 V / m

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1000 V / m

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10^{4} V / m

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50 V / m

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Solution

The correct option is D $10^{4} V / m$
We know $E = \frac{V}{d}$
$\Rightarrow E = \frac{100}{1 \times 10^{- 2}}$
$\Rightarrow E = 10^{4} v / m$

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The potential difference between two parallel plates 1 cm apart is 100V. The electric field strength between them is :

The correct option is D 104V/mWe know E=Vd⇒E=1001×10−2⇒E=104v/m

The correct option is D $10^{4} V / m$
We know $E = \frac{V}{d}$
$\Rightarrow E = \frac{100}{1 \times 10^{- 2}}$
$\Rightarrow E = 10^{4} v / m$