Answer:
a)4500V
b)750V
Step-by-step explanation:
Given:
Distance between the plate=
6.00 cm
We need to convert to m
Then the Distance between the plate=
0.06m
electric field strength between two parallel plates =
7.50 × 104 V/m .
Then E= 7.50 × 104 V/m .
(a) What is the potential difference between the plates?
potential difference between the plates can be calculated using the formula below
Δ Vab=ED
Where E is the given electric field strength
D= The Distance between the plate
ΔVab=7.50 × 10⁴V/m ×
0.06m
= 4500V
(b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 6.00 cm from the other?
the potential 1cm from the zero volt plate
Then the 1cm must be converted to m
= 0.01m
Let us say plate A as the plate at 0 volts:
The potential increases linearly going from plate A (0 V) to plate B (4500V).
Therefore,if the potential difference between A and B, separated by 6 cm, is 4500 V, then the potential difference between A and a point located at 1 cm from A is can be calculated also
If the plate with Lowest potential is taken to be zero then
=ΔVab=Vab-Vb=Va-0=Va=ED
Va=7.50 × 10⁴V/m × 0.01=750V