Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Certainly! Let's break down the problem step-by-step to find the resultant force on sphere C when it is exactly placed midway between spheres A and B.
### Step 1: Understand Initial Conditions
- Each sphere [tex]\(A\)[/tex] and [tex]\(B\)[/tex] has a charge of [tex]\(+q\)[/tex] coulomb.
- The force of repulsion between sphere [tex]\(A\)[/tex] and sphere [tex]\(B\)[/tex] is [tex]\( F = 10^{-5} \)[/tex] N.
- Coulomb’s constant, [tex]\( k = 8.99 \times 10^9 \)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex].
### Step 2: Calculate Distance [tex]\(r\)[/tex] between Spheres A and B
The force between two charges is given by Coulomb’s law:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Here, [tex]\( q_A = q_B = q \)[/tex], and substituting the given values:
[tex]\[ 10^{-5} = 8.99 \times 10^9 \frac{q^2}{r^2} \][/tex]
Solving for [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = 8.99 \times 10^9 \frac{q^2}{10^{-5}} \][/tex]
[tex]\[ r^2 = 8.99 \times 10^{14} q^2 \][/tex]
### Step 3: Touching Sphere C with A and B
When sphere [tex]\(C\)[/tex] is touched with [tex]\(A\)[/tex]:
- Charge on [tex]\(A\)[/tex] is [tex]\( q \)[/tex].
- Charge on [tex]\(C\)[/tex] becomes [tex]\( \frac{q + 0}{2} = \frac{q}{2} \)[/tex].
When sphere [tex]\(C\)[/tex] is then touched with [tex]\(B\)[/tex]:
- Charge on [tex]\(B\)[/tex] is [tex]\( q \)[/tex].
- Charge on [tex]\(C\)[/tex] after touching [tex]\(B\)[/tex] becomes [tex]\( \frac{\frac{q}{2} + q}{2} = \frac{3q}{4} \)[/tex].
Thus, the charge on sphere [tex]\(C\)[/tex] after touching both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is [tex]\( \frac{3q}{4} \)[/tex].
### Step 4: Place Sphere C Midway Between A and B
Placing sphere [tex]\(C\)[/tex] midway between [tex]\(A\)[/tex] and [tex]\(B\)[/tex] means the distance from [tex]\(A\)[/tex] to [tex]\(C\)[/tex] is [tex]\( \frac{r}{2} \)[/tex] and similarly for [tex]\(B\)[/tex] to [tex]\(C\)[/tex].
### Step 5: Calculate Resultant Forces
We calculate the forces on [tex]\(C\)[/tex] due to [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
1. Force between A and C:
- Distance is [tex]\( \frac{r}{2} \)[/tex].
- Charges are [tex]\( q_A = q \)[/tex] and [tex]\( q_C = \frac{3q}{4} \)[/tex].
Using Coulomb’s law:
[tex]\[ F_{A\text{-}C} = k \frac{q \cdot \frac{3q}{4}}{\left(\frac{r}{2}\right)^2} \][/tex]
Simplifying:
[tex]\[ F_{A\text{-}C} = k \frac{3q^2}{4} \cdot \frac{4}{r^2} \][/tex]
[tex]\[ F_{A\text{-}C} = 3 \frac{k q^2}{r^2} \][/tex]
Given [tex]\( \frac{k q^2}{r^2} = 10^{-5} \)[/tex]:
[tex]\[ F_{A\text{-}C} = 3 \times 10^{-5} N \][/tex]
2. Force between B and C:
- By symmetry, it is the same as the force between A and C because [tex]\(B\)[/tex] has the same charge as [tex]\(A\)[/tex] and the same distance to [tex]\(C\)[/tex].
[tex]\[ F_{B\text{-}C} = 3 \times 10^{-5} N \][/tex]
### Step 6: Resultant Force on Sphere C
Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are placed symmetrically on either side of [tex]\(C\)[/tex] and their forces are equal in magnitude but in opposite directions, they will cancel each other out.
Thus, the resultant force on sphere [tex]\(C\)[/tex] is:
[tex]\[ F_{\text{resultant on C}} = 0 N \][/tex]
Finally, we have determined the charges on spheres and calculated the forces. The resultant force on sphere C, placed exactly midway between A and B, is zero due to the symmetrical cancellation of forces.
### Step 1: Understand Initial Conditions
- Each sphere [tex]\(A\)[/tex] and [tex]\(B\)[/tex] has a charge of [tex]\(+q\)[/tex] coulomb.
- The force of repulsion between sphere [tex]\(A\)[/tex] and sphere [tex]\(B\)[/tex] is [tex]\( F = 10^{-5} \)[/tex] N.
- Coulomb’s constant, [tex]\( k = 8.99 \times 10^9 \)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex].
### Step 2: Calculate Distance [tex]\(r\)[/tex] between Spheres A and B
The force between two charges is given by Coulomb’s law:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Here, [tex]\( q_A = q_B = q \)[/tex], and substituting the given values:
[tex]\[ 10^{-5} = 8.99 \times 10^9 \frac{q^2}{r^2} \][/tex]
Solving for [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = 8.99 \times 10^9 \frac{q^2}{10^{-5}} \][/tex]
[tex]\[ r^2 = 8.99 \times 10^{14} q^2 \][/tex]
### Step 3: Touching Sphere C with A and B
When sphere [tex]\(C\)[/tex] is touched with [tex]\(A\)[/tex]:
- Charge on [tex]\(A\)[/tex] is [tex]\( q \)[/tex].
- Charge on [tex]\(C\)[/tex] becomes [tex]\( \frac{q + 0}{2} = \frac{q}{2} \)[/tex].
When sphere [tex]\(C\)[/tex] is then touched with [tex]\(B\)[/tex]:
- Charge on [tex]\(B\)[/tex] is [tex]\( q \)[/tex].
- Charge on [tex]\(C\)[/tex] after touching [tex]\(B\)[/tex] becomes [tex]\( \frac{\frac{q}{2} + q}{2} = \frac{3q}{4} \)[/tex].
Thus, the charge on sphere [tex]\(C\)[/tex] after touching both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is [tex]\( \frac{3q}{4} \)[/tex].
### Step 4: Place Sphere C Midway Between A and B
Placing sphere [tex]\(C\)[/tex] midway between [tex]\(A\)[/tex] and [tex]\(B\)[/tex] means the distance from [tex]\(A\)[/tex] to [tex]\(C\)[/tex] is [tex]\( \frac{r}{2} \)[/tex] and similarly for [tex]\(B\)[/tex] to [tex]\(C\)[/tex].
### Step 5: Calculate Resultant Forces
We calculate the forces on [tex]\(C\)[/tex] due to [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
1. Force between A and C:
- Distance is [tex]\( \frac{r}{2} \)[/tex].
- Charges are [tex]\( q_A = q \)[/tex] and [tex]\( q_C = \frac{3q}{4} \)[/tex].
Using Coulomb’s law:
[tex]\[ F_{A\text{-}C} = k \frac{q \cdot \frac{3q}{4}}{\left(\frac{r}{2}\right)^2} \][/tex]
Simplifying:
[tex]\[ F_{A\text{-}C} = k \frac{3q^2}{4} \cdot \frac{4}{r^2} \][/tex]
[tex]\[ F_{A\text{-}C} = 3 \frac{k q^2}{r^2} \][/tex]
Given [tex]\( \frac{k q^2}{r^2} = 10^{-5} \)[/tex]:
[tex]\[ F_{A\text{-}C} = 3 \times 10^{-5} N \][/tex]
2. Force between B and C:
- By symmetry, it is the same as the force between A and C because [tex]\(B\)[/tex] has the same charge as [tex]\(A\)[/tex] and the same distance to [tex]\(C\)[/tex].
[tex]\[ F_{B\text{-}C} = 3 \times 10^{-5} N \][/tex]
### Step 6: Resultant Force on Sphere C
Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are placed symmetrically on either side of [tex]\(C\)[/tex] and their forces are equal in magnitude but in opposite directions, they will cancel each other out.
Thus, the resultant force on sphere [tex]\(C\)[/tex] is:
[tex]\[ F_{\text{resultant on C}} = 0 N \][/tex]
Finally, we have determined the charges on spheres and calculated the forces. The resultant force on sphere C, placed exactly midway between A and B, is zero due to the symmetrical cancellation of forces.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
