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Two parallel plates of area [tex]$7.34 \cdot 10^{-4} \, \text{m}^2$[/tex] have [tex]$5.83 \cdot 10^{-8} \, \text{C}$[/tex] of charge placed on them. A [tex][tex]$6.62 \cdot 10^{-6} \, \text{C}$[/tex][/tex] charge [tex]$q_1$[/tex] is placed between the plates.

What is the magnitude of the electric force on [tex]$q_1$[/tex]?

Hint: How is force related to the field?

[tex] \text{Answer: } [?] \, \text{N} [/tex]

Sagot :

GinnyAnswer
Sure, let's go through the problem step-by-step to find the magnitude of the electric force on the charge [tex]\( q_1 \)[/tex].

### Step 1: Determine the Surface Charge Density (σ)
First, we need to calculate the surface charge density [tex]\( \sigma \)[/tex] on the plates. The surface charge density is defined as the charge per unit area:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]

Given:
[tex]\[ Q = 5.83 \times 10^{-8} \, \text{C} \][/tex]
[tex]\[ A = 7.34 \times 10^{-4} \, \text{m}^2 \][/tex]

So,
[tex]\[ \sigma = \frac{5.83 \times 10^{-8} \, \text{C}}{7.34 \times 10^{-4} \, \text{m}^2} \][/tex]

When you perform the division, you get:
[tex]\[ \sigma \approx 7.942779291553134 \times 10^{-5} \, \text{C/m}^2 \][/tex]

### Step 2: Determine the Electric Field (E) between the Plates
Next, we calculate the electric field [tex]\( E \)[/tex] between the plates. The electric field [tex]\( E \)[/tex] between parallel plates with surface charge density [tex]\( \sigma \)[/tex] is given by:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]

Where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space (a constant):
[tex]\[ \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \][/tex]

Substituting the calculated value of [tex]\( \sigma \)[/tex]:
[tex]\[ E = \frac{7.942779291553134 \times 10^{-5} \, \text{C/m}^2}{8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)} \][/tex]

This gives:
[tex]\[ E \approx 8970837.239160983 \, \text{N/C} \][/tex]

### Step 3: Calculate the Electric Force (F) on [tex]\( q_1 \)[/tex]
Lastly, we determine the force [tex]\( F \)[/tex] on the charge [tex]\( q_1 \)[/tex]. The electric force experienced by charge [tex]\( q_1 \)[/tex] in an electric field [tex]\( E \)[/tex] is given by:
[tex]\[ F = q_1 \cdot E \][/tex]

Given:
[tex]\[ q_1 = 6.62 \times 10^{-6} \, \text{C} \][/tex]

Substituting the values:
[tex]\[ F = 6.62 \times 10^{-6} \, \text{C} \times 8970837.239160983 \, \text{N/C} \][/tex]

This gives:
[tex]\[ F \approx 59.38694252324571 \, \text{N} \][/tex]

### Final Answer
The magnitude of the electric force on [tex]\( q_1 \)[/tex] is:
[tex]\[ 59.38694252324571 \, \text{N} \][/tex]
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