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2 years ago

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Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

two 5.5 cm × 5.5 cm electrodes.

2 answers:

Lera25 [3.4K]2 years ago

3 0

Electric field between the plates of parallel plate capacitor is given as

$E = \frac{Q}{A\epsilon_0}$

here area of plates of capacitor is given as

$A = 0.055 * 0.055$

$A = 3.025 * 10^{-3}$

also the maximum field strength is given as

$E = 3 * 10^6 N/C$

now we will plug in all data to find the maximum possible charge on capacitor plates

$3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}$

$Q = 8.03 * 10^{-8} C$

so the maximum charge that plate will hold will be given by above

victus00 [196]2 years ago

3 0

<h2>Complete Question:</h2>

Air "breaks down" when the electric field strength reaches 3×10⁶N/C, causing a spark. A parallel-plate capacitor is made from two 5.5cm × 5.5cm plates.

How many electrons must be transferred from one disk to the other to create a spark between the disks?

<h2>Answer:</h2>

5.02 x 10¹¹

<h2>Explanation:</h2>

The electric field E of a parallel-plate capacitor is given by;

E = Q / (A ε₀) --------------------------(i)

Where;

Q = the charge on the plates

A = Area of any of the plates

ε₀ = electric constant = 8.85 x 10⁻¹²F/m

<em>From the question;</em>

E = 3 x 10⁶N/C

A = 5.5cm x 5.5cm = 30.25cm² = 0.003025m²

<em>Substitute these values into equation (i) as follows;</em>

3 x 10⁶ = Q / (0.003025 x 8.85 x 10⁻¹²)

<em>Solve for Q;</em>

Q = 3 x 10⁶ x 0.003025 x 8.85 x 10⁻¹²

Q = 8.03 x 10⁻⁸

The charge on the capacitor is therefore, 8.03 x 10⁻⁸C

Now, to get the number of electrons that must be transferred from one disk to the other to create a spark between the disks, we use the relation;

Q = N x E ----------------------(ii)

Where;

Q = the charge = 8.03 x 10⁻⁸C [as calculated above]

N = the number of electrons

E = the charge of one electron = 1.6 x 10⁻¹⁹C [a known value]

<em>Substitute these values into equation (ii) as follows;</em>

8.03 x 10⁻⁸ = N x 1.6 x 10⁻¹⁹

<em>Solve for N;</em>

N = [8.03 x 10⁻⁸] / [1.6 x 10⁻¹⁹]

N = 5.02 x 10¹¹

Therefore, the number of electrons that must be transferred from one disk to the other to create a spark between the disks is 5.02 x 10¹³

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Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

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Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

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