A uniform electric field in positive `x`-direction on exists in space. An uncharged solid conducting sphere is now placed in region of this uniform electric field. The centre of this sphere of radius `R` lies at origin. Then electric potential at any point on `y-z` plane (i.e. `x=0` plane) due to only induced charged on surface of sphere.

Options

A. will be positive
B. will be negative
C. will be zero
D. may be positive, zero or negative

Explanation

To solve the problem, we need to analyze the situation where a uniform electric field is present, and an uncharged solid conducting sphere is placed in that field. The goal is to find the electric potential at any point on the y-z plane (where x=0) due to the induced charges on the surface of the sphere.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a uniform electric field directed in the positive x-direction.
- A solid conducting sphere of radius R is placed at the origin (0, 0, 0) of the coordinate system.

2. **Induced Charges on the Sphere**:
- When the conducting sphere is placed in the electric field, it will become polarized. The electric field will induce a positive charge on the hemisphere of the sphere that is facing the direction of the electric field (the right side, in this case) and a negative charge on the hemisphere that is facing away from the electric field (the left side).
- This results in a distribution of induced charges on the surface of the sphere.

3. **Electric Potential Calculation**:
- The electric potential at any point in space due to a charge distribution is given by the formula:
\[ V = k \sum \frac{q_i}{r_i} \]
where \( k \) is the Coulomb's constant, \( q_i \) is the charge, and \( r_i \) is the distance from the charge to the point where the potential is being calculated.
- However, in this case, we are interested in the potential at points on the y-z plane (where x=0) due to the induced charges.

4. **Symmetry Argument**:
- For any point on the y-z plane, there will be a corresponding point on the opposite side of the sphere that will have an equal and opposite charge.
- The potential due to the positive charge on the right hemisphere will be canceled out by the potential due to the negative charge on the left hemisphere at any point on the y-z plane.

5. **Conclusion**:
- Since the contributions to the potential from the positive and negative charges cancel each other out at every point on the y-z plane, the net electric potential at any point on the y-z plane due to the induced charges on the surface of the sphere is zero.

Thus, the electric potential at any point on the y-z plane due to the induced charges on the surface of the sphere is:
\[ \text{Electric Potential} = 0 \]