A flat square sheet of charge (side 50 cm) carries a uniform surface charge density. An electrons0.5 cm form a point near the center of the sheet experiences a force of `1.8xx10^(-12)N` directed away from the sheet. Determine the total charge on the sheet.

Explanation

To determine the total charge on the square sheet, we will follow these steps:

### Step 1: Understand the relationship between electric field (E), surface charge density (σ), and total charge (Q).
The electric field (E) due to an infinite sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where:
- \( \sigma \) is the surface charge density,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 2: Relate surface charge density to total charge.
The surface charge density (σ) can also be expressed in terms of total charge (Q) and area (A):
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the area of the sheet.

### Step 3: Calculate the area of the square sheet.
Given that the side of the square sheet is 50 cm, we convert this to meters:
\[ \text{Side} = 0.5 \, \text{m} \]
Thus, the area \( A \) is:
\[ A = \text{Side}^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \]

### Step 4: Substitute the expression for σ into the electric field formula.
Substituting \( \sigma \) into the electric field formula gives:
\[ E = \frac{Q}{2 \epsilon_0 A} \]

### Step 5: Relate the force experienced by the electron to the electric field.
The force \( F \) experienced by a charge \( e \) (the charge of the electron) in an electric field \( E \) is given by:
\[ F = eE \]
where \( e = 1.6 \times 10^{-19} \, \text{C} \).

### Step 6: Substitute the expression for E into the force equation.
We can substitute the expression for \( E \) into the force equation:
\[ F = e \left(\frac{Q}{2 \epsilon_0 A}\right) \]

### Step 7: Rearrange the equation to solve for total charge Q.
Rearranging the equation for \( Q \) gives:
\[ Q = \frac{2 \epsilon_0 A F}{e} \]

### Step 8: Substitute the known values into the equation.
Now we can substitute the known values:
- \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( A = 0.25 \, \text{m}^2 \)
- \( F = 1.8 \times 10^{-12} \, \text{N} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)

Substituting these values:
\[ Q = \frac{2 \times (8.854 \times 10^{-12}) \times (0.25) \times (1.8 \times 10^{-12})}{1.6 \times 10^{-19}} \]

### Step 9: Calculate the total charge Q.
Calculating this gives:
\[ Q \approx \frac{2 \times 8.854 \times 10^{-12} \times 0.25 \times 1.8 \times 10^{-12}}{1.6 \times 10^{-19}} \]
\[ Q \approx -5.0 \times 10^{-5} \, \text{C} \]
or
\[ Q \approx -50 \, \mu\text{C} \]

### Final Answer:
The total charge on the sheet is approximately \( -50 \, \mu\text{C} \).

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