Question

A parallel-plate capacitor is formed from two 2.7 cm -diameter electrodes spaced 1.4 mm apart. The electric field strength inside the capacitor is 6.0 x106 N/C

Answer 1

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 $8.4\times10^{3}\ V$