Question

2 point charges 2nc and -2nc are placed 0.1m apart. find the resultant electric intensity at a point at distance 0.1m from each charge.

Answer 1

Given:

2 point charges 2nc and -2nc are placed 0.1m apart.

To Find:

The resultant electric intensity at a point at distance 0.1m from each charge.

Solution:

Let A and B be the points where, 2nC and -2nC are kept.

Let C and D be points at 0.1 m distance away from this dipole.

Let C be close to A and D be close to B.

Electric field at C =

• Electric field due to 2nC + Electric field due to -2nC.
• Electric field at C = $\frac{1}{4\pi\epsilon_0}$ Q/R² = $\frac{1}{4\pi\epsilon_0}$2 x $10^{-9}$ / (0.1)² + $\frac{1}{4\pi\epsilon_0}$ -2 x $10^{-9}$/(0.2)²

• Electric field at C = 9x$10^{9}$ x 2 x $10^{-9}$/ (0.1)² +  9x$10^{9}$ x -2 x $10^{-9}$/ (0.2)²
• Electric field at C = 100 x 18 - 25x18 = 1350N/C

Electric field at D =

• Electric field due to 2nC + Electric field due to -2nC.

• Electric field at D = $\frac{1}{4\pi\epsilon_0}$ Q/R² = $\frac{1}{4\pi\epsilon_0}$2 x $10^{-9}$ / (0.2)² + $\frac{1}{4\pi\epsilon_0}$ -2 x $10^{-9}$/(0.1)²

• Electric field at D = 9x$10^{9}$ x 2 x $10^{-9}$/ (0.2)² +  9x$10^{9}$ x -2 x $10^{-9}$/ (0.1)²

• Electric field at D = 25 x 18 - 100x18 = -1350N/C

The resultant electric intensity at a point at distance 0.1m from each charge is 1350N/C in magnitude, but direction of field is opposite to each other.