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A set of charged plates have an area of [tex]$5.10 \times 10^{-3} \, \text{m}^2$[/tex] and a separation of [tex]$1.42 \times 10^{-5} \, \text{m}$[/tex]. How much charge must be placed on the plates to create a potential difference of [tex][tex]$125 \, \text{V}$[/tex][/tex] across them?

(The answer is [tex] \qquad \times 10^{-7} \, \text{C}[/tex]. Just fill in the number, not the power.)


Sagot :

Certainly! Let's solve this problem step-by-step.

### Step-by-Step Solution

1. Given Values:
- Area [tex]\( A = 5 \times 10^{-3} \)[/tex] square meters ([tex]\( m^2 \)[/tex])
- Separation [tex]\( d = 1.42 \times 10^{-5} \)[/tex] meters ([tex]\( m \)[/tex])
- Voltage [tex]\( V = 125 \)[/tex] Volts ([tex]\( V \)[/tex])

2. Constants:
- Permittivity of free space [tex]\( \epsilon_0 = 8.854 \times 10^{-12} \)[/tex] Farads per meter ([tex]\( F/m \)[/tex])

3. Calculating the Capacitance:

The formula for the capacitance [tex]\( C \)[/tex] of a parallel plate capacitor is given by:
[tex]\[ C = \epsilon_0 \frac{A}{d} \][/tex]
Plugging in the values:
[tex]\[ C = (8.854 \times 10^{-12}) \times \frac{5 \times 10^{-3}}{1.42 \times 10^{-5}} \][/tex]
Upon calculating this:
[tex]\[ C \approx 3.1176 \times 10^{-9} \, \text{Farads (F)} \][/tex]

4. Calculating the Charge:

The charge [tex]\( Q \)[/tex] stored in the capacitor is given by:
[tex]\[ Q = C \times V \][/tex]
Plugging in the values:
[tex]\[ Q = (3.1176 \times 10^{-9}) \times 125 \][/tex]
Upon calculating this:
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{Coulombs (C)} \][/tex]

5. Converting the Charge to the Required Format:

To express the charge [tex]\( Q \)[/tex] in the form [tex]\( * \times 10^{-7} \)[/tex] Coulombs (C):
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{C} \][/tex]
Thus, the charge required is simply [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs [tex]\( (10^{-7} \, \text{C}) \)[/tex].

So, the answer is:

[tex]\[ 3.8970 \][/tex]

Therefore, the charge must be [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs to create a potential difference of 125 V across the plates.