Question

Two-point charges of 2 micro C but opposite in sign are placed 10 cm apart to calculate the electric field at a. Point 10 cm from mid point on the axial line of dipole.

Answer 1

Step 1, Given data and related image

Charge = 2 micro C

Distance = 10 cm

Given situation can be represented as

Step 2, Finding the electric field

We can write,

$E_{net}=\frac{k{q}_1}{r^2}+\frac{k{q}_2}{r^2}$

$Or,E_{net}=k[\frac{2}{{(5\times {10}^2)}^2}-\frac{2}{{(15\times {10}^2)}^2}]$

Or,

$E_{net}=\frac{16\times {10}^7}{25}=0.64\times {10}^7$

N/C

Hence the electric field is $0.64\times {10}^7$ N/C