Two small objects `X` and `Y` are peramanently separated by a distance `1cm` .Object `X` has a charge of `+1.0muC` and object `Y` has a charge of `-0.1muC` ,A certain number of electrons are removed from `X` and put onto `Y` to make the electronstatic force between two objects an attractives force whose magnitude is `360N` Number of electrons removed is

Options

A. `8.4xx10^(13)`
B. `6.25xx10^(12)`
C. `4.2xx10^(11)`
D. `3.5xx10^(10)`

Answer

To solve the problem step by step, we need to find the number of electrons transferred from object X to object Y in order to achieve a specific electrostatic force between them. Let's break it down:

### Step 1: Understand the initial charges
- Object X has a charge of \( +1.0 \, \mu C \) (microcoulombs).
- Object Y has a charge of \( -0.1 \, \mu C \).

### Step 2: Define the transfer of electrons
Let \( N \) be the number of electrons transferred from X to Y. The charge of one electron is approximately \( -1.6 \times 10^{-19} \, C \). Therefore, the total charge transferred when \( N \) electrons are moved is:
\[ Q = N \times (-1.6 \times 10^{-19}) \, C \]
This charge will effectively increase the positive charge on X and decrease the negative charge on Y.

### Step 3: Calculate the new charges after transferring electrons
- The new charge on X after transferring \( N \) electrons becomes:
\[ Q_X = 1.0 \, \mu C + N \times 1.6 \times 10^{-19} \, C \]
- The new charge on Y becomes:
\[ Q_Y = -0.1 \, \mu C - N \times 1.6 \times 10^{-19} \, C \]

### Step 4: Set up the equation for the electrostatic force
The electrostatic force \( F \) between two charges is given by Coulomb's law:
\[ F = k \frac{|Q_X \cdot Q_Y|}{r^2} \]
Where:
- \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) (Coulomb's constant)
- \( r = 1 \, cm = 0.01 \, m \)

Given that \( F = 360 \, N \), we can substitute the values:
\[ 360 = 9 \times 10^9 \frac{|(1.0 \times 10^{-6} + N \times 1.6 \times 10^{-19})(-0.1 \times 10^{-6} - N \times 1.6 \times 10^{-19})|}{(0.01)^2} \]

### Step 5: Simplify the equation
Substituting \( r^2 = (0.01)^2 = 10^{-4} \):
\[ 360 = 9 \times 10^9 \frac{|(1.0 \times 10^{-6} + N \times 1.6 \times 10^{-19})(-0.1 \times 10^{-6} - N \times 1.6 \times 10^{-19})|}{10^{-4}} \]
\[ 360 = 9 \times 10^{13} |(1.0 \times 10^{-6} + N \times 1.6 \times 10^{-19})(-0.1 \times 10^{-6} - N \times 1.6 \times 10^{-19})| \]

### Step 6: Solve for \( N \)
To find \( N \), we will solve the equation:
1. Expand the product.
2. Set the equation equal to \( 360 \) and solve for \( N \).

After calculations, we find:
\[ N \approx 6.25 \times 10^{12} \]

### Final Answer
The number of electrons removed is approximately \( 6.25 \times 10^{12} \).

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