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: Find the surface charge density first.
The field between the plates depends only on how densely the charge is packed on a plate, through \(E = \sigma/\varepsilon_0\). So begin by extracting the surface charge density \(\sigma\):
\[\sigma = \varepsilon_0 E\]

Step 2: Put in the numbers.
\[\sigma = (8.85\times10^{-12})\times(100) = 8.85\times10^{-10}\ \text{C/m}^2\]

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

Step 4: Charge = density \(\times\) area.
Since \(\sigma = Q/A\), the total charge on one plate is
\[Q = \sigma A = (8.85\times10^{-10})\times(1\times10^{-2})\]

Step 5: Compute.
\[Q = 8.85\times10^{-12}\ \text{C}\]
The same value is obtained, confirming the result. Each plate holds equal and opposite charge of this magnitude.
\[\boxed{Q \approx 8.85\times10^{-12}\ \text{C} \ (8.85\ \text{pC})}\]