9. Find the force between two small charges in the following cases: a) $$ -1.0 imes 10^{-7} \mathrm{C} $$ and $$ +2.0 imes 10^{-7} \mathrm{C} $$ b) $$ +4.0 imes 10^{7} \mathrm{C} $$ and $$ +3.0 imes 10^{-7} \mathrm{C} $$ placed 4 cm apart in vacuum. Find in each case the intensity at P midway of the two charges. $$ \left[\left(4 \pi \varepsilon_{0} ight)^{-1}=9.0 imes 10^{\circ} \mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2} ight] $$

Question

9. Find the force between two small charges in the following cases: a) $$ -1.0 	imes 10^{-7} \mathrm{C} $$ and $$ +2.0 	imes 10^{-7} \mathrm{C} $$ b) $$ +4.0 	imes 10^{7} \mathrm{C} $$ and $$ +3.0 	imes 10^{-7} \mathrm{C} $$ placed 4 cm apart in vacuum. Find in each case the intensity at P midway of the two charges. $$ \left[\left(4 \pi \varepsilon_{0}ight)^{-1}=9.0 	imes 10^{\circ} \mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2}ight] $$

UpStudy AI · Accepted Answer

To find the force between two small charges and the intensity at point P midway between them, we can use Coulomb's law and the formula for electric field intensity.

### Known Conditions:
1. Coulomb's Law: 
   $$
   F = k \frac{|q_1 q_2|}{r^2}
   $$
   where $$ F $$ is the force between the charges, $$ k = \frac{1}{4 \pi \varepsilon_0} $$, $$ q_1 $$ and $$ q_2 $$ are the charges, and $$ r $$ is the distance between them.

2. Electric Field Intensity (E) at point P:
   $$
   E = k \frac{|q|}{r^2}
   $$
   where $$ q $$ is the charge creating the field and $$ r $$ is the distance from the charge to point P.

### Given Data:
- $$ k = 9.0 \times 10^9 \, \mathrm{N m^2 C^{-2}} $$
- Distance between charges $$ r = 4 \, \mathrm{cm} = 0.04 \, \mathrm{m} $$
- Midway distance from each charge to point P $$ r_P = \frac{r}{2} = 0.02 \, \mathrm{m} $$

### Case (a): Charges $$ -1.0 \times 10^{-7} \mathrm{C} $$ and $$ +2.0 \times 10^{-7} \mathrm{C} $$

1. **Calculate the force $$ F $$**:
   $$
   F = k \frac{|q_1 q_2|}{r^2} = 9.0 \times 10^9 \frac{|(-1.0 \times 10^{-7})(2.0 \times 10^{-7})|}{(0.04)^2}
   $$

2. **Calculate the electric field intensity $$ E $$ at point P**:
   - Due to $$ q_1 $$:
   $$
   E_1 = k \frac{|q_1|}{r_P^2} = 9.0 \times 10^9 \frac{|(-1.0 \times 10^{-7})|}{(0.02)^2}
   $$
   - Due to $$ q_2 $$:
   $$
   E_2 = k \frac{|q_2|}{r_P^2} = 9.0 \times 10^9 \frac{|(2.0 \times 10^{-7})|}{(0.02)^2}
   $$

### Case (b): Charges $$ +4.0 \times 10^{-7} \mathrm{C} $$ and $$ +3.0 \times 10^{-7} \mathrm{C} $$

1. **Calculate the force $$ F $$**:
   $$
   F = k \frac{|q_1 q_2|}{r^2} = 9.0 \times 10^9 \frac{|(4.0 \times 10^{-7})(3.0 \times 10^{-7})|}{(0.04)^2}
   $$

2. **Calculate the electric field intensity $$ E $$ at point P**:
   - Due to $$ q_1 $$:
   $$
   E_1 = k \frac{|q_1|}{r_P^2} = 9.0 \times 10^9 \frac{|(4.0 \times 10^{-7})|}{(0.02)^2}
   $$
   - Due to $$ q_2 $$:
   $$
   E_2 = k \frac{|q_2|}{r_P^2} = 9.0 \times 10^9 \frac{|(3.0 \times 10^{-7})|}{(0.02)^2}
   $$

Now, let's perform the calculations for both cases.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(3\times 10-7\right)}{0.02^{2}}$$
- step1: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(30-7\right)}{0.02^{2}}$$
- step2: Subtract the numbers:
  $$\frac{9\times 10^{9}\times 23}{0.02^{2}}$$
- step3: Convert the expressions:
  $$\frac{9\times 10^{9}\times 23}{\left(\frac{1}{50}\right)^{2}}$$
- step4: Multiply the terms:
  $$\frac{207\times 10^{9}}{\left(\frac{1}{50}\right)^{2}}$$
- step5: Simplify the expression:
  $$\frac{207\times 10^{9}}{\frac{1}{50^{2}}}$$
- step6: Rewrite the expression:
  $$517500\times 10^{9}$$
Calculate or simplify the expression $$ 9.0 * 10^9 * (1.0 * 10^-7 * 2.0 * 10^-7) / (0.04)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(1\times 10-7\times 2\times 10-7\right)}{0.04^{2}}$$
- step1: Calculate:
  $$\frac{9\times 10^{9}\left(10-7\times 2\times 10-7\right)}{0.04^{2}}$$
- step2: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(10-140-7\right)}{0.04^{2}}$$
- step3: Subtract the numbers:
  $$\frac{9\times 10^{9}\left(-137\right)}{0.04^{2}}$$
- step4: Convert the expressions:
  $$\frac{9\times 10^{9}\left(-137\right)}{\left(\frac{1}{25}\right)^{2}}$$
- step5: Multiply the numbers:
  $$\frac{-1233\times 10^{9}}{\left(\frac{1}{25}\right)^{2}}$$
- step6: Rewrite the fraction:
  $$-\frac{1233\times 10^{9}}{\left(\frac{1}{25}\right)^{2}}$$
- step7: Simplify the expression:
  $$-\frac{1233\times 10^{9}}{\frac{1}{25^{2}}}$$
- step8: Rewrite the expression:
  $$-770625\times 10^{9}$$
Calculate or simplify the expression $$ 9.0 * 10^9 * (2.0 * 10^-7) / (0.02)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(2\times 10-7\right)}{0.02^{2}}$$
- step1: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(20-7\right)}{0.02^{2}}$$
- step2: Subtract the numbers:
  $$\frac{9\times 10^{9}\times 13}{0.02^{2}}$$
- step3: Convert the expressions:
  $$\frac{9\times 10^{9}\times 13}{\left(\frac{1}{50}\right)^{2}}$$
- step4: Multiply the terms:
  $$\frac{117\times 10^{9}}{\left(\frac{1}{50}\right)^{2}}$$
- step5: Simplify the expression:
  $$\frac{117\times 10^{9}}{\frac{1}{50^{2}}}$$
- step6: Rewrite the expression:
  $$292500\times 10^{9}$$
Calculate or simplify the expression $$ 9.0 * 10^9 * (4.0 * 10^-7 * 3.0 * 10^-7) / (0.04)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(4\times 10-7\times 3\times 10-7\right)}{0.04^{2}}$$
- step1: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(40-7\times 3\times 10-7\right)}{0.04^{2}}$$
- step2: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(40-210-7\right)}{0.04^{2}}$$
- step3: Subtract the numbers:
  $$\frac{9\times 10^{9}\left(-177\right)}{0.04^{2}}$$
- step4: Convert the expressions:
  $$\frac{9\times 10^{9}\left(-177\right)}{\left(\frac{1}{25}\right)^{2}}$$
- step5: Multiply the numbers:
  $$\frac{-1593\times 10^{9}}{\left(\frac{1}{25}\right)^{2}}$$
- step6: Rewrite the fraction:
  $$-\frac{1593\times 10^{9}}{\left(\frac{1}{25}\right)^{2}}$$
- step7: Simplify the expression:
  $$-\frac{1593\times 10^{9}}{\frac{1}{25^{2}}}$$
- step8: Rewrite the expression:
  $$-995625\times 10^{9}$$
Calculate or simplify the expression $$ 9.0 * 10^9 * (1.0 * 10^-7) / (0.02)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(1\times 10-7\right)}{0.02^{2}}$$
- step1: Calculate:
  $$\frac{9\times 10^{9}\left(10-7\right)}{0.02^{2}}$$
- step2: Subtract the numbers:
  $$\frac{9\times 10^{9}\times 3}{0.02^{2}}$$
- step3: Convert the expressions:
  $$\frac{9\times 10^{9}\times 3}{\left(\frac{1}{50}\right)^{2}}$$
- step4: Multiply the terms:
  $$\frac{27\times 10^{9}}{\left(\frac{1}{50}\right)^{2}}$$
- step5: Simplify the expression:
  $$\frac{27\times 10^{9}}{\frac{1}{50^{2}}}$$
- step6: Rewrite the expression:
  $$67500\times 10^{9}$$
Calculate or simplify the expression $$ 9.0 * 10^9 * (4.0 * 10^-7) / (0.02)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{9\times 10^{9}\left(4\times 10-7\right)}{0.02^{2}}$$
- step1: Multiply the numbers:
  $$\frac{9\times 10^{9}\left(40-7\right)}{0.02^{2}}$$
- step2: Subtract the numbers:
  $$\frac{9\times 10^{9}\times 33}{0.02^{2}}$$
- step3: Convert the expressions:
  $$\frac{9\times 10^{9}\times 33}{\left(\frac{1}{50}\right)^{2}}$$
- step4: Multiply the terms:
  $$\frac{297\times 10^{9}}{\left(\frac{1}{50}\right)^{2}}$$
- step5: Simplify the expression:
  $$\frac{297\times 10^{9}}{\frac{1}{50^{2}}}$$
- step6: Rewrite the expression:
  $$742500\times 10^{9}$$
### Results of the Calculations

#### Case (a): Charges $$ -1.0 \times 10^{-7} \mathrm{C} $$ and $$ +2.0 \times 10^{-7} \mathrm{C} $$

1. **Force $$ F $$**:
   $$
   F = -7.70625 \times 10^{14} \, \mathrm{N} \quad (\text{attractive force due to opposite charges})
   $$

2. **Electric Field Intensity $$ E $$ at point P**:
   - Due to $$ q_1 $$:
   $$
   E_1 = 6.75 \times 10^{13} \, \mathrm{N/C}
   $$
   - Due to $$ q_2 $$:
   $$
   E_2 = 2.925 \times 10^{14} \, \mathrm{N/C}
   $$

Since $$ q_1 $$ is negative and $$ q_2 $$ is positive, the electric fields will point in opposite directions. The net electric field intensity $$ E $$ at point P will be:
   $$
   E = E_2 - E_1 = 2.925 \times 10^{14} - 6.75 \times 10^{13} = 2.25 \times 10^{14} \, \mathrm{N/C}
   $$

#### Case (b): Charges $$ +4.0 \times 10^{-7} \mathrm{C} $$ and $$ +3.0 \times 10^{-7} \mathrm{C} $$

1. **Force $$ F $$**:
   $$
   F = -9.95625 \times 10^{14} \, \mathrm{N} \quad (\text{repulsive force due to like charges})
   $$

2. **Electric Field Intensity $$ E $$ at point P**:
   - Due to $$ q_1 $$:
   $$
   E_1 = 7.425 \times 10^{14} \, \mathrm{N/C}
   $$
   - Due to $$ q_2 $$:
   $$
   E_2 = 5.175 \times 10^{14} \, \mathrm{N/C}
   $$

Since both charges are positive, the electric fields will point in the same direction. The net electric field intensity $$ E $$ at point P will be:
   $$
   E = E_1 + E_2 = 7.425 \times 10^{14} + 5.175 \times 10^{14} = 1.26 \times 10^{15} \, \mathrm{N/C}
   $$

### Summary of Results
- **Case (a)**:
  - Force: $$ -7.70625 \times 10^{14} \, \mathrm{N} $$
  - Electric Field Intensity at P: $$ 2.25 \times 10^{14} \, \mathrm{N/C} $$

- **Case (b)**:
  - Force: $$ -9.95625 \times 10^{14} \, \mathrm{N} $$
  - Electric Field Intensity at P: $$ 1.26 \times 10^{15} \, \mathrm{N/C} $$
- **Case (a)**: - Force: $$ -7.70625 \times 10^{14} \, \mathrm{N} $$ (attractive) - Electric Field Intensity at P: $$ 2.25 \times 10^{14} \, \mathrm{N/C} $$ - **Case (b)**: - Force: $$ -9.95625 \times 10^{14} \, \mathrm{N} $$ (repulsive) - Electric Field Intensity at P: $$ 1.26 \times 10^{15} \, \mathrm{N/C} $$

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