Sure, I'll guide you through the steps necessary to solve for the charge [tex]\( q \)[/tex] given the conditions of the problem.
### Given:
1. Energy used to move the charge: [tex]\( 10 \text{ mJ} \)[/tex]
2. Initial position [tex]\( x_1 = 25 \text{ cm} \)[/tex]
3. Final position [tex]\( x_2 = -25 \text{ cm} \)[/tex]
4. Electric field [tex]\( E = -20 \text{ N/C} \)[/tex]
### Step-by-Step Solution:
#### Step 1: Convert All Given Values to SI Units
1. Energy:
The energy given is [tex]\( 10 \text{ mJ} \)[/tex]. Convert it to joules:
[tex]\[
10 \text{ mJ} = 10 \times 10^{-3} \text{ J} = 0.01 \text{ J}
\][/tex]
2. Positions:
Convert the positions from centimeters to meters:
[tex]\[
x_1 = 25 \text{ cm} = 25 \times 10^{-2} \text{ m} = 0.25 \text{ m}
\][/tex]
[tex]\[
x_2 = -25 \text{ cm} = -25 \times 10^{-2} \text{ m} = -0.25 \text{ m}
\][/tex]
3. Electric Field:
The electric field is already in SI units ([tex]\(\text{N/C}\)[/tex]), so no conversion is necessary:
[tex]\[
E = -20 \text{ N/C}
\][/tex]
#### Step 2: Calculate the Potential Difference
The potential difference ([tex]\(\Delta V\)[/tex]) is given by:
[tex]\[
\Delta V = E \cdot \Delta x
\][/tex]
where [tex]\(\Delta x = x_2 - x_1\)[/tex]. Substitute the values:
[tex]\[
\Delta x = -0.25 \text{ m} - 0.25 \text{ m} = -0.5 \text{ m}
\][/tex]
Thus,
[tex]\[
\Delta V = (-20 \text{ N/C}) \times (-0.5 \text{ m}) = 10 \text{ V}
\][/tex]
#### Step 3: Calculate the Charge [tex]\( q \)[/tex]
Using the energy equation involving electric potential:
[tex]\[
\text{Energy} = q \cdot \Delta V
\][/tex]
Rearrange to solve for [tex]\( q \)[/tex]:
[tex]\[
q = \frac{\text{Energy}}{\Delta V}
\][/tex]
Substitute the values:
[tex]\[
q = \frac{0.01 \text{ J}}{10 \text{ V}} = 0.001 \text{ C}
\][/tex]
### Final Answer:
- The potential difference (Voltage) is [tex]\( 10 \text{ V} \)[/tex].
- The charge [tex]\( q \)[/tex] is [tex]\( 0.001 \text{ C} \)[/tex].
