Answer :
To determine the work done when an electron is moved from one plate to another, we need to consider the electric field, the distance between the plates, and the charge of the electron. Here's the detailed step-by-step solution:
1. Convert the distance to meters:
The distance between the plates is given as \(3.0 \, \text{cm}\). To use this in our calculations correctly, we need to convert it to meters:
[tex]\[ 3.0 \, \text{cm} = 3.0 \times 10^{-2} \, \text{m} = 0.03 \, \text{m} \][/tex]
2. Identify the given values:
- Electric field (\(E\)) = \(700 \, \text{N/C}\)
- Distance between plates (\(d\)) = \(0.03 \, \text{m}\)
- Charge of an electron (\(q\)) = \(-1.6 \times 10^{-19} \, \text{C}\) (Note: The negative sign indicates the charge of the electron, but since we are calculating work done, we'll consider the magnitude for this context.)
3. Calculate the work done:
The work done (\(W\)) in moving a charge in an electric field is given by the formula:
[tex]\[ W = E \cdot q \cdot d \][/tex]
Here,
\( E = 700 \, \text{N/C} \)
\( q = 1.6 \times 10^{-19} \, \text{C} \)
\( d = 0.03 \, \text{m} \)
Substituting these values in:
[tex]\[ W = 700 \times 1.6 \times 10^{-19} \times 0.03 \][/tex]
4. Compute the numerical value:
[tex]\[ W = 700 \times 1.6 \times 0.03 \times 10^{-19} \][/tex]
Calculate the product of the constants first:
[tex]\[ 700 \times 1.6 = 1120 \][/tex]
Then:
[tex]\[ 1120 \times 0.03 = 33.6 \][/tex]
Now apply the exponent:
[tex]\[ 33.6 \times 10^{-19} = 3.36 \times 10^{-18} \, \text{J} \][/tex]
Thus, the work done when an electron is moved from the positive to the negative plate is \(3.36 \times 10^{-18} \, \text{J}\). The correct answer is:
[tex]\[ \boxed{3.36 \times 10^{-18} \, \text{J}} \][/tex]
1. Convert the distance to meters:
The distance between the plates is given as \(3.0 \, \text{cm}\). To use this in our calculations correctly, we need to convert it to meters:
[tex]\[ 3.0 \, \text{cm} = 3.0 \times 10^{-2} \, \text{m} = 0.03 \, \text{m} \][/tex]
2. Identify the given values:
- Electric field (\(E\)) = \(700 \, \text{N/C}\)
- Distance between plates (\(d\)) = \(0.03 \, \text{m}\)
- Charge of an electron (\(q\)) = \(-1.6 \times 10^{-19} \, \text{C}\) (Note: The negative sign indicates the charge of the electron, but since we are calculating work done, we'll consider the magnitude for this context.)
3. Calculate the work done:
The work done (\(W\)) in moving a charge in an electric field is given by the formula:
[tex]\[ W = E \cdot q \cdot d \][/tex]
Here,
\( E = 700 \, \text{N/C} \)
\( q = 1.6 \times 10^{-19} \, \text{C} \)
\( d = 0.03 \, \text{m} \)
Substituting these values in:
[tex]\[ W = 700 \times 1.6 \times 10^{-19} \times 0.03 \][/tex]
4. Compute the numerical value:
[tex]\[ W = 700 \times 1.6 \times 0.03 \times 10^{-19} \][/tex]
Calculate the product of the constants first:
[tex]\[ 700 \times 1.6 = 1120 \][/tex]
Then:
[tex]\[ 1120 \times 0.03 = 33.6 \][/tex]
Now apply the exponent:
[tex]\[ 33.6 \times 10^{-19} = 3.36 \times 10^{-18} \, \text{J} \][/tex]
Thus, the work done when an electron is moved from the positive to the negative plate is \(3.36 \times 10^{-18} \, \text{J}\). The correct answer is:
[tex]\[ \boxed{3.36 \times 10^{-18} \, \text{J}} \][/tex]
