Question

If an electric generator produces a potential difference of \(180\,{kV}\) in a spark of length \(9\,{cm}\), then the maximum electric field generated is:

• A. \( 3 \times 10^6 \, {N/C} \)
• B. \( 10^6 \, {N/C} \)
• C. \( 2 \times 10^6 \, {N/C} \)
• D. \( 4 \times 10^6 \, {N/C} \)

Hint:

Use \( E = \frac{V}{d} \) for uniform electric fields, ensuring units of distance are in meters.

Answer 1

The Correct Option is C

Electric field \( E \) is calculated as: \[ E = \frac{V}{d} \] Given: \( V = 180\,{kV} = 1.8 \times 10^5 \, {V} \), \( d = 9\,{cm} = 0.09\,{m} \) \[ E = \frac{1.8 \times 10^5}{0.09} = 2 \times 10^6 \, {N/C} \]