A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0 × 106 N/C. What is the charge (in nC) on each electrode?

To find the charge on each electrode of the parallel-plate capacitor, we can use the relationship between the electric field \( E \), the charge \( Q \), the area \( A \) of the plates, and the permittivity of free space \( \varepsilon_0 \):

\[
E = \frac{Q}{A \cdot \varepsilon_0}
\]

Where:
- \( E \) is the electric field strength (1.0 × 10^6 N/C).
- \( A \) is the area of one of the electrodes.
- \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

First, we need to calculate the area \( A \) of one electrode, which is circular:

\[
A = \pi r^2
\]

The radius \( r \) can be found from the diameter:

\[
r = \frac{6.0 \, \text{cm}}{2} = 3.0 \, \text{cm} = 0.03 \, \text{m}
\]

Now we calculate the area:

\[
A = \pi (0.03 \, \text{m})^2 \approx 2.827 \times 10^{-3} \, \text{m}^2
\]

Now we can rearrange the equation for charge \( Q \):

\[
Q = E \cdot A \cdot \varepsilon_0
\]

Substituting the known values:

\[
Q = (1.0 \times 10^6 \, \text{N/C}) \cdot (2.827 \times 10^{-3} \, \text{m}^2) \cdot (8.85 \times 10^{-12} \, \text{F/m})
\]

Calculating \( Q \):

\[
Q \approx (1.0 \times 10^6) \cdot (2.827 \times 10^{-3}) \cdot (8.85 \times 10^{-12}) \approx 2.5 \times 10^{-18} \, \text{C}
\]

Now converting this charge into nanocoulombs (nC):

\[
Q \approx 2.5 \times 10^{-18} \, \text{C} \times 10^{9} \, \text{nC/C} \approx 2.5 \times 10^{-9} \, \text{nC} \approx 2.5 \, \text{nC}
\]

Thus, the charge on each electrode is approximately **2.5 nC**.

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. How does the capacitance change if the distance between the plates is increased?
2. What factors influence the electric field strength in a capacitor?
3. How would the charge on the plates change if the diameter of the electrodes was doubled?
4. What is the relationship between capacitance, voltage, and charge in a capacitor?
5. How does introducing a dielectric material affect the charge and electric field in the capacitor? 

**Tip:** Remember to always keep track of units when performing calculations in physics!

A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0 × 10^6 N/C. What is the charge (in nC) on each electrode?

Discover how to calculate the charge on each electrode of a parallel-plate capacitor with 6.0 cm diameter and 2.0 mm spacing, given an electric field strength of 1.0 × 10^6 N/C. The final charge calculated is approximately 2.5 nC.

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