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3 years ago

6

A parallel-plate capacitor is formed from two 2.7 cm -diameter electrodes spaced 1.4 mm apart. The electric field strength insid

e the capacitor is 6.0 x106 N/C

1 answer:

Andre45 [30]3 years ago

3 0

Answer:

The potential difference between the plates is $8.4\times10^{3}\ V$

Explanation:

Given that,

Distance = 1.4 mm

Electric field strength $E= 6.0\times10^{6}\ N/C$

Let the potential difference is V.

We need to calculate the potential difference between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$V=Ed$

Where, V = potential

d = distance

Put the value into the formula

$V=6.0\times10^{6}\times1.4\times10^{-3}$

$V=8.4\times10^{3}\ V$

Hence, The potential difference between the plates is $8.4\times10^{3}\ V$

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