An isolated conducting sphere has a 20 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a current of 1.000 000 0 A out of it. How long would it take for the sphere to increase in potential by 1000 V?

Answer

To solve the problem, we need to determine how long it will take for the potential of an isolated conducting sphere to increase by 1000 V, given the currents flowing into and out of the sphere.

### Step-by-Step Solution

1. **Identify the Given Values:**
- Radius of the sphere, \( R = 20 \, \text{cm} = 0.20 \, \text{m} \)
- Current flowing into the sphere, \( I_{\text{in}} = 1.0000020 \, \text{A} \)
- Current flowing out of the sphere, \( I_{\text{out}} = 1.0000000 \, \text{A} \)
- Change in potential, \( \Delta V = 1000 \, \text{V} \)

2. **Calculate the Net Current:**
The net current \( I \) flowing into the sphere is given by:
\[ I = I_{\text{in}} - I_{\text{out}} = 1.0000020 \, \text{A} - 1.0000000 \, \text{A} = 0.0000020 \, \text{A} \]

3. **Relate Potential Change to Charge:**
The potential \( V \) of a charged sphere is given by:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
where \( Q \) is the charge on the sphere and \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)).

4. **Express Change in Charge:**
The change in charge \( \Delta Q \) can be expressed in terms of the current and time:
\[ \Delta Q = I \Delta t \]

5. **Relate Change in Potential to Change in Charge:**
The change in potential \( \Delta V \) is related to the change in charge:
\[ \Delta V = \frac{\Delta Q}{4 \pi \epsilon_0 R} \]
Substituting for \( \Delta Q \):
\[ \Delta V = \frac{I \Delta t}{4 \pi \epsilon_0 R} \]

6. **Rearranging to Solve for Time:**
Rearranging the equation to solve for \( \Delta t \):
\[ \Delta t = \frac{4 \pi \epsilon_0 R \Delta V}{I} \]

7. **Substituting the Values:**
Now we substitute the known values into the equation:
\[ \Delta t = \frac{4 \pi (8.85 \times 10^{-12} \, \text{F/m}) (0.20 \, \text{m}) (1000 \, \text{V})}{0.0000020 \, \text{A}} \]

8. **Calculating the Result:**
Performing the calculations:
\[ \Delta t = \frac{4 \pi (8.85 \times 10^{-12}) (0.20) (1000)}{0.0000020} \]
\[ \Delta t \approx \frac{4 \times 3.14 \times 8.85 \times 0.20 \times 1000}{0.0000020} \]
\[ \Delta t \approx \frac{22.24 \times 10^{-9}}{0.0000020} \approx 1.112 \times 10^{-2} \, \text{s} \approx 0.01112 \, \text{s} \]

### Final Answer:
The time it would take for the sphere to increase in potential by 1000 V is approximately \( 0.01112 \, \text{s} \).

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