Question
Two large horizontal and parallel metal plates are placed 2.0 cm apart in air and the upper plate is maintained at a positive potential relative to the lower plate so that the field strength between them is
500
V
m
-
1
.
What is the potential difference between the plates is?
Question 6Select one:
10 V
100 V
1000 V
0.1 V
500
V
m
-
1
.
What is the potential difference between the plates is?
Question 6Select one:
10 V
100 V
1000 V
0.1 V
Answers
GPT-4o mini
The electric field strength (\(E\)) between the plates is given as \(500 \, \text{V/m}\), and the distance (\(d\)) between the plates is \(2.0 \, \text{cm}\), which is \(0.02 \, \text{m}\).
The potential difference (\(V\)) between the plates can be calculated using the formula:
\[
V = E \cdot d
\]
Plugging in the values:
\[
V = 500 \, \text{V/m} \cdot 0.02 \, \text{m} = 10 \, \text{V}
\]
Thus, the potential difference between the plates is \(10 \, \text{V}\).
The correct answer is:
**10 V**
The potential difference (\(V\)) between the plates can be calculated using the formula:
\[
V = E \cdot d
\]
Plugging in the values:
\[
V = 500 \, \text{V/m} \cdot 0.02 \, \text{m} = 10 \, \text{V}
\]
Thus, the potential difference between the plates is \(10 \, \text{V}\).
The correct answer is:
**10 V**