Question

Two infinitely long parallel plates of equal areas, $6$ cm$^2$, are separated by a distance of $1$ cm. While one of the plates has a charge of $+10$ nC and the other has $-10$ nC. The magnitude of the electric field between the plates, if $\varepsilon_0 = \dfrac{10^{-9}{36\pi}$ F/m, is}

• A. $0.6\pi$ kV/m
• B. $6\pi$ kV/m
• C. $600\pi$ kV/m
• D. $60\pi$ V/m
• E. $6\pi$ V/m

Hint:

For parallel plates, $E = \sigma/\varepsilon_0$ and always convert cm$^2$ to m$^2$.

Answer 1

The Correct Option is B

Concept:
Electric field between parallel plates: \[ E = \frac{\sigma}{\varepsilon_0} \]

Step 1: Convert area.
\[ A = 6 \text{ cm}^2 = 6 \times 10^{-4} \text{ m}^2 \]

Step 2: Find surface charge density.
\[ \sigma = \frac{Q}{A} = \frac{10 \times 10^{-9}}{6 \times 10^{-4}} = \frac{10^{-5}}{6} \]

Step 3: Substitute in formula.
\[ E = \frac{10^{-5}/6}{10^{-9}/36\pi} \]

Step 4: Simplify.
\[ E = \frac{10^{-5}}{6} \times \frac{36\pi}{10^{-9}} = 6\pi \times 10^4 \text{ V/m} \] \[ = 6\pi \text{ kV/m} \]