A point charge of magnitude `+1 mu C` is fixed at `(0,0,0).` An isolated uncharged spherical conductor, is fixed with its center at (4, 0, 0) cm. I'he potential and the induced electric field at the centre of the sphere is :

Options

A. `1.8 xx 10^5 V and - 5.625 xx 10^6 V//m `
B. `0 V and 0 V // m`
C. `2.25 xx 10^5 V and - 5.625 xx 10^6 V//m`
D. `2.25 xx 10^5 V and 0 V//m`

Answer

To solve the problem, we need to calculate the potential and the induced electric field at the center of an uncharged spherical conductor due to a point charge of +1 µC located at the origin (0, 0, 0). The center of the sphere is located at (4, 0, 0) cm.

### Step 1: Calculate the Distance
First, we need to find the distance \( r \) from the point charge to the center of the sphere. The coordinates of the point charge are (0, 0, 0) and the center of the sphere is at (4, 0, 0).

\[ r = \text{Distance} = |4 - 0| = 4 \text{ cm} = 4 \times 10^{-2} \text{ m} \]

### Step 2: Calculate the Potential at the Center of the Sphere
The potential \( V \) due to a point charge is given by the formula:

\[ V = k \frac{q}{r} \]

where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant),
- \( q = 1 \times 10^{-6} \, \text{C} \) (the charge),
- \( r = 4 \times 10^{-2} \, \text{m} \) (the distance).

Substituting the values:

\[ V = 9 \times 10^9 \frac{1 \times 10^{-6}}{4 \times 10^{-2}} = 9 \times 10^9 \times \frac{1}{4} \times 10^2 \]

Calculating this gives:

\[ V = \frac{9}{4} \times 10^9 \times 10^2 = 2.25 \times 10^5 \, \text{V} \]

### Step 3: Calculate the Induced Electric Field at the Center of the Sphere
The electric field \( E \) due to a point charge is given by the formula:

\[ E = k \frac{q}{r^2} \]

Using the same values for \( k \), \( q \), and \( r \):

\[ E = 9 \times 10^9 \frac{1 \times 10^{-6}}{(4 \times 10^{-2})^2} \]

Calculating \( (4 \times 10^{-2})^2 = 16 \times 10^{-4} = 1.6 \times 10^{-3} \):

\[ E = 9 \times 10^9 \frac{1 \times 10^{-6}}{1.6 \times 10^{-3}} = 9 \times 10^9 \times \frac{1}{1.6} \times 10^{-3} \]

Calculating this gives:

\[ E = \frac{9}{1.6} \times 10^6 = 5.625 \times 10^6 \, \text{V/m} \]

### Step 4: Determine the Induced Electric Field Inside the Conductor
Inside a conductor, the electric field is zero. Since the sphere is uncharged and isolated, the induced electric field at the center of the sphere is:

\[ E_{\text{induced}} = -5.625 \times 10^6 \, \text{V/m} \]

### Final Answers
- The potential at the center of the sphere is \( 2.25 \times 10^5 \, \text{V} \).
- The induced electric field at the center of the sphere is \( -5.625 \times 10^6 \, \text{V/m} \).