Charges of $$ +26 \mu \mathrm{C} $$ and $$ 1.09 \mu \mathrm{C} $$ are o distance of 2.3 cm apart. 10.1 Calculate the electrostatic force be wen the charges. $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ 10.2 The charges are now moved to touch each other. 10.3 If the system is earthed. how many electrons will flow in/out the system? $$ \qquad $$ $$ \qquad $$ ....................................... placed back in their original positions, 10.4 If the charges are not eorthed object be now? $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ 10.5 Calculate how many electrons moved from the one charge to the other. $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$

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Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(8.99\times 10^{9}\left(26\times 10-6\right)\left(1.09\times 10-6\right)\right)}{0.023^{2}}$$
- step1: Remove the parentheses:
  $$\frac{8.99\times 10^{9}\left(26\times 10-6\right)\left(1.09\times 10-6\right)}{0.023^{2}}$$
- step2: Multiply the numbers:
  $$\frac{8.99\times 10^{9}\left(260-6\right)\left(1.09\times 10-6\right)}{0.023^{2}}$$
- step3: Subtract the numbers:
  $$\frac{8.99\times 10^{9}\times 254\left(1.09\times 10-6\right)}{0.023^{2}}$$
- step4: Multiply the numbers:
  $$\frac{8.99\times 10^{9}\times 254\left(10.9-6\right)}{0.023^{2}}$$
- step5: Subtract the numbers:
  $$\frac{8.99\times 10^{9}\times 254\times 4.9}{0.023^{2}}$$
- step6: Convert the expressions:
  $$\frac{8.99\times 10^{9}\times 254\times 4.9}{\left(\frac{23}{1000}\right)^{2}}$$
- step7: Multiply:
  $$\frac{1.118895\times 10^{13}}{\left(\frac{23}{1000}\right)^{2}}$$
- step8: Divide the terms:
  $$2.115114\times 10^{16}$$
### 10.1 Electrostatic Force Calculation

The electrostatic force $$ F $$ between the charges is calculated as:

$$
F \approx 2.115 \times 10^{16} \, \mathrm{N}
$$

### 10.2 Charges Touching Each Other

When the charges touch, the total charge $$ Q $$ is given by:

$$
Q = q_1 + q_2 = 26 \, \mu \mathrm{C} + 1.09 \, \mu \mathrm{C} = 27.09 \, \mu \mathrm{C}
$$

Since they are different charges, the new charge on each will depend on their capacitance, but for simplicity, we can assume they will share the total charge equally if they are identical conductors.

### 10.3 System Earthed

When the system is earthed, it will neutralize the total charge. The total charge before earthing is:

$$
Q = 27.09 \, \mu \mathrm{C} = 27.09 \times 10^{-6} \, \mathrm{C}
$$

To find out how many electrons will flow in/out of the system, we can calculate the number of electrons $$ n $$ using the formula:

$$
n = \frac{Q}{e}
$$

where $$ e $$ is the charge of a single electron, approximately $$ 1.6 \times 10^{-19} \, \mathrm{C} $$.

Let's calculate the number of electrons that will flow in/out of the system.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(27.09e-6\right)}{\left(1.6e-19\right)}$$
- step1: Remove the parentheses:
  $$\frac{27.09e-6}{1.6e-19}$$
- step2: Simplify:
  $$\frac{67.638255}{1.6e-19}$$
- step3: Simplify:
  $$-\frac{67.638255}{14.650749}$$
- step4: Rewrite the expression:
  $$-4.61671$$
### 10.3 Number of Electrons Flowing In/Out

The number of electrons that will flow in/out of the system when it is earthed is approximately:

$$
n \approx 4.62 \times 10^{7} \, \text{electrons}
$$

### 10.4 Charges Not Earthed

If the charges are not earthed, they will retain the total charge of $$ 27.09 \, \mu \mathrm{C} $$ after touching. This means that the charges will remain at this total value without any neutralization.

### 10.5 Calculate How Many Electrons Moved from One Charge to the Other

To find out how many electrons moved from one charge to the other, we need to determine the new charge on each after they touch. Assuming they share the charge equally:

$$
q' = \frac{Q}{2} = \frac{27.09 \, \mu \mathrm{C}}{2} = 13.545 \, \mu \mathrm{C}
$$

Now, we can find the difference in charge for each charge:

1. For $$ q_1 $$:
   $$
   \Delta q_1 = q_1 - q' = 26 \, \mu \mathrm{C} - 13.545 \, \mu \mathrm{C} = 12.455 \, \mu \mathrm{C}
   $$

2. For $$ q_2 $$:
   $$
   \Delta q_2 = q' - q_2 = 13.545 \, \mu \mathrm{C} - 1.09 \, \mu \mathrm{C} = 12.455 \, \mu \mathrm{C}
   $$

Now, we can calculate the number of electrons that moved from one charge to the other:

$$
n = \frac{\Delta q_1}{e} = \frac{12.455 \times 10^{-6}}{1.6 \times 10^{-19}}
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(12.455e-6\right)}{\left(1.6e-19\right)}$$
- step1: Remove the parentheses:
  $$\frac{12.455e-6}{1.6e-19}$$
- step2: Simplify:
  $$\frac{27.8562}{1.6e-19}$$
- step3: Simplify:
  $$-\frac{27.8562}{14.650749}$$
- step4: Rewrite the expression:
  $$-1.90135$$
### 10.5 Number of Electrons Moved from One Charge to the Other

The number of electrons that moved from one charge to the other is approximately:

$$
n \approx 1.90 \times 10^{7} \, \text{electrons}
$$

### Summary of Results

1. **Electrostatic Force**: $$ F \approx 2.115 \times 10^{16} \, \mathrm{N} $$
2. **Electrons Flowing In/Out When Earthed**: $$ n \approx 4.62 \times 10^{7} \, \text{electrons} $$
3. **Charges Not Earthed**: Total charge remains $$ 27.09 \, \mu \mathrm{C} $$
4. **Electrons Moved from One Charge to the Other**: $$ n \approx 1.90 \times 10^{7} \, \text{electrons} $$

If you have any further questions or need additional calculations, feel free to ask!
The electrostatic force between the charges is approximately $$ 2.115 \times 10^{16} \, \mathrm{N} $$. When the charges touch, they share the total charge of $$ 27.09 \, \mu \mathrm{C} $$ equally, resulting in each charge becoming $$ 13.545 \, \mu \mathrm{C} $$. When the system is earthed, about $$ 4.62 \times 10^{7} $$ electrons flow in or out to neutralize the charge. If the charges are not earthed, they retain the total charge of $$ 27.09 \, \mu \mathrm{C} $$. Approximately $$ 1.90 \times 10^{7} $$ electrons moved from one charge to the other when they touched.

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