Question

you have a parallel - plate capacitor. part a determine the magnitude of the average e - field between the plates if a 160 - v potential difference exists across the plates. their separation is 0.50 cm. express your answer with the appropriate units. e = value units part b a spark will jump if the magnitude of the e - field exceeds 3.0×10^6 v/m when air separates the plates. what is the closest distance the plates can be placed to each other without sparking? express your answer with the appropriate units. d_min = value units

Explanation:

Step1: Recall electric - field formula

The formula for the electric field between two parallel - plates is $E=\frac{V}{d}$, where $E$ is the electric field, $V$ is the potential difference, and $d$ is the plate separation.

Step2: Solve for $E$ in Part A

Given $V = 160\ V$ and $d=0.50\ cm=0.005\ m$. Substitute into the formula $E=\frac{V}{d}=\frac{160\ V}{0.005\ m}=3.2\times 10^{4}\ V/m$.

Step3: Solve for $d$ in Part B

We know the maximum electric field $E_{max}=3.0\times 10^{6}\ V/m$ and $V = 160\ V$. Rearranging the formula $E=\frac{V}{d}$ to solve for $d$, we get $d=\frac{V}{E}$. Substitute $V = 160\ V$ and $E = 3.0\times 10^{6}\ V/m$ into the formula, $d=\frac{160\ V}{3.0\times 10^{6}\ V/m}\approx5.33\times 10^{-5}\ m = 53.3\ \mu m$.

Answer:

Part A: $E = 3.2\times 10^{4}\ V/m$
Part B: $d_{min}=5.33\times 10^{-5}\ m$ (or $53.3\ \mu m$)