A parallel plate capacitor has a capacitance of \(2.0 \mu \mathrm{F}\) and plate separation of \(1.0 \mathrm{mm} .\) (a) How much potential difference can be placed across the capacitor before dielectric breakdown of air occurs \(\left(E_{\max }=3 \times 10^{6} \mathrm{V} / \mathrm{m}\right) ?\) (b) What is the magnitude of the greatest charge the capacitor can store before breakdown?

To find the maximum potential difference, first, we need to determine the maximum electric field allowed before dielectric breakdown occurs. We are given the maximum electric field strength in the problem: $$ E_{\max} = 3 \times 10^6 V/m $$. Next, we can use the relationship between electric field, potential difference, and distance between the plates: $$ E = \frac{V}{d} $$, where \(E\) is the electric field, \(V\) is the potential difference, and \(d\) is the distance between the plates. Rearrange the equation to solve for \(V\): $$ V = E \times d $$. Finally, plug in the values for \(E_{\max}\) and \(d\) into the equation to find the maximum potential difference: $$ V_{\max} = (3 \times 10^6 V/m)(1 \times 10^{-3} m) = 3000 V. $$ So, the maximum potential difference is \(3000 V\).

To find the maximum charge stored in the capacitor, we will use the capacitance and the maximum potential difference. The formula relating these quantities is: $$ Q = C \times V $$, where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the potential difference. Plugging in the values for \(C\) and \(V_{\max}\) into the equation: $$ Q_{\max} = (2.0 \times 10^{-6} F)(3000 V) = 6 \times 10^{-3} C $$. The maximum charge that the capacitor can store before dielectric breakdown is \(6 \times 10^{-3} C\). In summary, we found that the maximum potential difference the capacitor can have before dielectric breakdown is \(3000 V\), and the maximum charge it can store before breakdown is \(6 \times 10^{-3} C\).