Answer :
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.
