Question

Four point charges of 1 pc, 2 pc, 3 pc,
and 4 pc, are placed along x axis at x = 1
m, x = 2 m, x = 3 m and x = 4 m
respectively. The electric field
intensity at origin due to first three
charges is​

Answer 1

Given :

Four point charges of charges :

1 pc , 2 pc , 3 pc , 4pc

Distances of these charges from origin along x axis :

1 m , 2 m , 3 m , 4 m

To Find :

The electric field intensity at origin due to  the first the charges = ?

Solution :

∴1 C = $10^{12}pc$

⇒$1pc = 10^{-12} C$

We know that formula of electric field is :

$E=\frac{kq}{r^{2} }$

So here by superposition theorem the electric field at origin due to the three charges will be equal to the sum of electric field due to the three charges individually t the origin.

⇒E= $E_{1} +E_{2} + E_{3}$

$E =K( \frac{q_{1}}{r_{1}^{2} } +\frac{q_{2}}{r_{2}^{2} } +\frac{q_{3}}{r_{3}^{2} } )$

$E= 9\times 10^{9} ( \frac{1}{1^{2} } +\frac{2}{2^{2} } +\frac{3}{3^{2} } )\times 10^{-12} \frac{N}{C}$

$E=9\times 10^{-3} \times \frac{11}{6} \\\\E=16.5\times 10^{-3} \frac{N}{C}$

So the electric field due to the three charges is 0.165 N per C or Volt per meter .