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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

2 Answers

2 votes

Answer : Electric potential,
V=1.8* 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=(V)/(d)


V=E* d


V= 875\ N/C* 0.002\ m


V=1.75\ V

or


V=1.8* 10^0\ V

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

Hence, this is the required solution.

User SDJSK
by
7.3k points
4 votes
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 \, (N)/(C))*(0.002 \, m) \\\\ = 1.75 \, (J)/(C) \\\\ =1.75 \, V

Answer: 1.8 V (nearest tenth)

User Shady Boshra
by
7.2k points