Two point charges +4q and +q are placed 30cm apart. At what point on the line joining them is the electric field zero

Options

A. 15 cm from charge q
B. 20 cm from charge 4 q
C. 7.5 cm from charge q
D. 5cm from charge q

Explanation

To find the point where the electric field is zero between two point charges +4q and +q separated by a distance of 30 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let the charge +4q be at point A (0 cm) and charge +q be at point B (30 cm).
- The distance between the charges is given as 30 cm.

2. **Set Up the Problem**:
- We need to find a point P on the line joining the two charges where the electric field is zero.
- Let the distance from charge +4q to point P be x cm. Therefore, the distance from charge +q to point P will be (30 - x) cm.

3. **Understand Electric Field Direction**:
- The electric field due to a positive charge is directed away from the charge.
- At point P, the electric field due to charge +q will be directed away from +q (to the left) and the electric field due to charge +4q will be directed away from +4q (to the right).

4. **Write the Expression for Electric Fields**:
- The electric field \( E_q \) due to charge +q at point P is given by:
\[ E_q = \frac{k \cdot q}{(30 - x)^2} \]
- The electric field \( E_{4q} \) due to charge +4q at point P is given by:
\[ E_{4q} = \frac{k \cdot 4q}{x^2} \]

5. **Set the Electric Fields Equal**:
- For the electric field to be zero at point P, the magnitudes of the electric fields due to both charges must be equal:
\[ E_q = E_{4q} \]
- Therefore, we can write:
\[ \frac{k \cdot q}{(30 - x)^2} = \frac{k \cdot 4q}{x^2} \]

6. **Cancel Common Terms**:
- We can cancel \( k \) and \( q \) from both sides (assuming \( q \neq 0 \)):
\[ \frac{1}{(30 - x)^2} = \frac{4}{x^2} \]

7. **Cross Multiply**:
- Cross multiplying gives:
\[ x^2 = 4(30 - x)^2 \]

8. **Expand and Rearrange**:
- Expanding the right side:
\[ x^2 = 4(900 - 60x + x^2) \]
\[ x^2 = 3600 - 240x + 4x^2 \]
- Rearranging gives:
\[ 0 = 3x^2 - 240x + 3600 \]

9. **Solve the Quadratic Equation**:
- Dividing the entire equation by 3:
\[ x^2 - 80x + 1200 = 0 \]
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{80 \pm \sqrt{6400 - 4800}}{2} \]
\[ x = \frac{80 \pm \sqrt{1600}}{2} \]
\[ x = \frac{80 \pm 40}{2} \]
- This gives two solutions:
\[ x = 60 \text{ cm (not valid, outside the range)} \quad \text{and} \quad x = 20 \text{ cm} \]

10. **Conclusion**:
- The point where the electric field is zero is 20 cm from charge +4q.

### Final Answer:
The electric field is zero at a point 20 cm from charge +4q.