Question

the potential gradient between the two charged plates having separation of 0.5 volt and potential difference of 12 volts

Answer 1

Answer:

As we learned

Electric Field Intensity -

\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}

- wherein

E=\frac{V}{d}=\frac{10}{2\times 10^{-2}}=500\; N/C

Option 1)

20\; \; N/C

Option 2)

500\; \; N/C

Option 3)

5\; \; N/C

Option 4)

250\; \; N/C