Problem: Two point charges of 5nC and 20nC are separated by a distance of 1.5m. Find the point on the line joining them at which the net electric field is zero.

Solution:

Introduction: In this problem, we are given two point charges and we need to find the point on the line joining them where the net electric field is zero. To solve this problem, we will use the concept of electric field due to point charge and superposition principle.

Electric Field due to Point Charge: The electric field due to a point charge at a distance r from the charge is given by:
E = kq/r^2
where E is the electric field, k is the Coulomb's constant, q is the charge and r is the distance between the charge and the point where we want to calculate the electric field.

Superposition Principle: The electric field at any point due to a system of charges is equal to the vector sum of the electric fields due to individual charges at that point.

Calculation: Let us assume that the point on the line joining the two charges where the net electric field is zero is x meters away from the 5nC charge. Then, the distance between the 20nC charge and the point will be (1.5 - x) meters.

Using the electric field due to point charge formula, we can calculate the electric field due to each charge at the point x.

Electric field due to 5nC charge at point x:
E1 = kq1/x^2

Electric field due to 20nC charge at (1.5 - x) distance from point x:
E2 = kq2/(1.5 - x)^2

According to the superposition principle, the net electric field at point x is given by the vector sum of E1 and E2.
E = E1 + E2

Since we want the net electric field to be zero, we can set E1 equal to -E2 and solve for x.

kq1/x^2 = -kq2/(1.5 - x)^2

Solving for x, we get:
x = 0.875m

Therefore, the point on the line joining the two charges where the net electric field is zero is 0.875m away from the 5nC charge.

Conclusion: In this problem, we used the electric field due to point charge formula and superposition principle to calculate the point on the line joining two charges where the net electric field is zero. The final answer was 0.875m away from the 5nC charge.