Question

The area of each plate of a parallel plate condenser is 100 cm² and the intensity of the electric field between the two plates is 100 newton/coulomb. How much charge is there on each plate?
(Take \( \varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N·m}^2 \).)

Hint:

Use \( E = \sigma/\varepsilon_0 = Q/(A\varepsilon_0) \), so \( Q = \varepsilon_0 E A \); remember to convert 100 cm² to \( 10^{-2}\ \text{m}^2 \).

Answer 1

Step 1: Relevant formula.
For a parallel plate capacitor the uniform electric field between the plates is related to the charge on a plate by
\[ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\,\varepsilon_0} \]
where \( \sigma = Q/A \) is the surface charge density, \( Q \) the charge on each plate and \( A \) the area of each plate.

Step 2: Rearrange for the charge.
\[ Q = \varepsilon_0\, E\, A \]

Step 3: Convert the area to SI units.
\[ A = 100\ \text{cm}^2 = 100\times10^{-4}\ \text{m}^2 = 1\times10^{-2}\ \text{m}^2 \]

Step 4: Substitute the values.
\[ Q = (8.85\times10^{-12})\times(100)\times(1\times10^{-2}) \]

Step 5: Arithmetic.
\[ Q = 8.85\times10^{-12}\times(100\times10^{-2}) = 8.85\times10^{-12}\times1 \]
\[ Q = 8.85\times10^{-12}\ \text{C} \]

So each plate carries a charge of about \( 8.85\times10^{-12}\ \text{C} \) (about 8.85 pC), one plate positive and the other negative.
\[\boxed{Q \approx 8.85\times10^{-12}\ \text{C}}\]