1.The electric field strength between two parallel plates separated by 6.00 cm is 7.50 × 104 V/m . (a) What is the potential difference between the plates? (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?

a)4500V

b)750V

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

answer: strength i guess

explanation:

a=d/t

a="triangle"v/t

explanation:

a = acceleration