Question

The electric field between two parallel plates has a magnitude of 875 N/C. The positive plate is 0.002 m away from the negative plate. What is the electric potential difference between the plates? 2.3 × 10-6 V 1.8 × 100 V 8.8 × 102 V 4.4 × 104 V

Answer 1

Given:
E = 875 N/C electric field
d = 00.002 m, distance between parallel plates

Note that
1 V = 1 J/C

The electric potential difference is
$V=E*d \\\\ =(875 \, \frac{N}{C})*(0.002 \, m) \\\\ = 1.75 \, \frac{J}{C} \\\\ =1.75 \, V$

Answer: 1.8 V (nearest tenth)

Answer 2

Answer : Electric potential, $V=1.8\times 10^0\ V$

Explanation :

It is given that,

Magnitude of electric field, $E=875\ N/C$

Distance between two plates, $d=0.002\ m$

The relation between the electric field and the electric potential is given by :

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

$V=E\times d$

$V= 875\ N/C\times 0.002\ m$

$V=1.75\ V$

or

$V=1.8\times 10^0\ V$

So, the magnitude of electric potential is given by option (2).

Hence, this is the required solution.