Two charge +aq and -q are kept apart. Then , at any point on the right bisector of line joining the two charge

Options

A. the electric field strength is zero
B. the electric potential is zero
C. both electric potential and electric field strength are zero
D. both electric potential and electric field strength are non-zero

Explanation

To solve the problem of determining the electric potential and electric field at any point on the right bisector of the line joining two charges +aq and -q, we can follow these steps:

### Step 1: Understanding the Configuration
We have two charges: +aq located at point A and -q located at point B. The line joining these two charges is bisected at point O, which is equidistant from both charges. Let the distance from point O to each charge be 'a'.

### Step 2: Identifying the Point of Interest
We need to analyze a point P located on the right bisector of the line segment AB. This means that point P is directly perpendicular to the midpoint O of the line segment AB.

### Step 3: Calculating Electric Potential at Point P
The electric potential (V) due to a point charge is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r} \]
Where:
- \( Q \) is the charge,
- \( r \) is the distance from the charge to the point of interest,
- \( \epsilon_0 \) is the permittivity of free space.

For point P, the distance from charge +aq (at A) to P is \( r_{AP} \) and from charge -q (at B) to P is \( r_{BP} \).

### Step 4: Expressing Distances
Since point P is on the right bisector, the distances can be expressed as:
- \( r_{AP} = \sqrt{a^2 + d^2} \) (where d is the perpendicular distance from O to P)
- \( r_{BP} = \sqrt{a^2 + d^2} \)

### Step 5: Calculating Total Electric Potential at Point P
The total electric potential at point P due to both charges is:
\[ V_P = V_{A} + V_{B} \]
\[ V_P = \frac{1}{4\pi \epsilon_0} \cdot \frac{+aq}{\sqrt{a^2 + d^2}} + \frac{1}{4\pi \epsilon_0} \cdot \frac{-q}{\sqrt{a^2 + d^2}} \]

### Step 6: Simplifying the Expression
Combining the potentials:
\[ V_P = \frac{1}{4\pi \epsilon_0} \cdot \left( \frac{+aq - q}{\sqrt{a^2 + d^2}} \right) \]
\[ V_P = \frac{1}{4\pi \epsilon_0} \cdot \frac{(a-1)q}{\sqrt{a^2 + d^2}} \]

### Step 7: Determining Electric Field at Point P
The electric field (E) due to a point charge is given by:
\[ E = -\frac{dV}{dr} \]
However, since we are looking at the right bisector, the contributions to the electric field from both charges will cancel each other out due to symmetry.

### Conclusion
At any point on the right bisector of the line joining the two charges, the electric potential is non-zero (depends on the values of a and q), while the electric field strength is zero due to the symmetry of the configuration.

### Final Answer
- Electric Potential: Non-zero
- Electric Field Strength: Zero