 
15. The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50 x 10^4V/m. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

Let's break down the problem step by step.

We are given:
- Electric field strength \( E = 7.50 \times 10^4 \, \text{V/m} \)
- Separation between the plates \( d = 4.00 \, \text{cm} = 0.0400 \, \text{m} \)

### Part (a) – Potential difference between the plates:

The potential difference \( \Delta V \) between two parallel plates is related to the electric field \( E \) and the separation \( d \) by the equation:

\[
\Delta V = E \cdot d
\]

Substitute the values:

\[
\Delta V = (7.50 \times 10^4 \, \text{V/m}) \cdot (0.0400 \, \text{m})
\]

\[
\Delta V = 3000 \, \text{V}
\]

Thus, the potential difference between the plates is \( 3000 \, \text{V} \).

---

### Part (b) – Potential at 1.00 cm from the plate at 0 volts:

We are now asked to find the potential at a point 1.00 cm from the plate with the lowest potential (taken as 0 volts).

- Distance from the 0 V plate = \( x = 1.00 \, \text{cm} = 0.0100 \, \text{m} \)
- The distance from the other plate (higher potential plate) would be \( 3.00 \, \text{cm} = 0.0300 \, \text{m} \).

Since the electric field \( E \) between parallel plates is uniform, the potential \( V \) at a distance \( x \) from the zero-volts plate is given by:

\[
V = E \cdot x
\]

Substitute the values:

\[
V = (7.50 \times 10^4 \, \text{V/m}) \cdot (0.0100 \, \text{m})
\]

\[
V = 750 \, \text{V}
\]

Thus, the potential at 1.00 cm from the plate with 0 volts is \( 750 \, \text{V} \).

---

### Final Answers:
- **(a)** The potential difference between the plates is \( 3000 \, \text{V} \).
- **(b)** The potential at 1.00 cm from the plate with 0 V is \( 750 \, \text{V} \).

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Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:
1. What happens to the potential difference if the separation between the plates is doubled?
2. How would the electric field change if the potential difference between the plates increases?
3. How is the capacitance of the parallel plates related to the electric field and plate separation?
4. What is the energy stored between the plates if the plates form a capacitor?
5. How does the presence of a dielectric between the plates affect the potential difference and electric field?

**Tip**: Always keep track of units when calculating, especially when converting between centimeters and meters.

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50 x 10^4 V/m. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

Learn how to calculate the potential difference between parallel conducting plates separated by 4.00 cm with an electric field strength of 7.50 x 10^4 V/m. Includes step-by-step solution and explanation suitable for students in Grades 10-12.

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