Solution:

Given, charges Q = Q and 4Q are placed 30 cm apart.

To find: The point where the value of electric field will be zero.

Formula: The electric field due to a point charge Q at a distance r from it is given by

E = kQ/r²

Where k is the Coulomb’s constant (9 × 10^9 Nm²/C²).

The electric field at a point P due to two point charges Q1 and Q2 placed at a distance d from each other is given by

E = k(Q1/d² – Q2/d²)

Where d is the distance between the two charges.

Analysis:

The electric fields due to Q and 4Q at a point P at a distance x from Q and (30 – x) from 4Q are given by

E1 = kQ/x²

E2 = k(4Q)/(30 – x)²

The net electric field E at point P is given by

E = E1 + E2

Equating E to zero, we get

kQ/x² = k(4Q)/(30 – x)²

On simplification, we get

(30 – x)² = 4x²

900 – 60x + x² = 4x²

3x² + 60x – 900 = 0

x² + 20x – 300 = 0

On solving the above quadratic equation, we get

x = 10√13 – 10

The point where the electric field is zero is at a distance of 10√13 – 10 cm from the charge Q.

Conclusion:

Hence, the point where the value of electric field will be zero is at a distance of 10√13 – 10 cm from the charge Q.