Principle of superpostion

Principle of superpostion

This principle tells us that if charge Q is acted upon by several charges q1 , q2 …. qn , then the net force on Q can be found out by calculating separately the forces F1 , F2 … Fn , exerted by q1 , q2 , …. qn respectively on Q and then adding these forces vectorially.
Their resultant F is the total force on q due to the collection of charges.

Thus the net force on a charge Q is the vector sum of forces on Q due to all the remaining charges.

$\displaystyle \vec{F} = \vec{F_{1}} + \vec{F_{2}} +\vec{F_{3}} + ….. + \vec{F_{n}}$

Example : It is required to hold equal charges, q in equilibrium at the corners of a square. What charge when placed at the centre of the square will do this ?

Solution: Let ABCD be a square of side a , and Q be the charge placed at the centre.

$\displaystyle \vec{F_{BA}} = k \frac{q^2}{a^2}\hat{i}$

$\displaystyle \vec{F_{BC}} = -k \frac{q^2}{a^2}\hat{j}$

$\displaystyle \vec{F_{BD}} = k \frac{q^2}{(a\sqrt2)^2}(cos45\hat{i}-sin45\hat{j})$

$\displaystyle \vec{F_{BQ}} = k \frac{q Q}{(a/\sqrt2)^2}(cos45\hat{i}-sin45\hat{j})$

$\displaystyle F_x = k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ]$

$\displaystyle F_y = -k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ]$

Net force on the charge at B is

$\displaystyle \vec{F}= F_x\hat{i} + F_y\hat{j}$

For charge, q to be in equilibrium at B, the net force on it must be zero.

$\displaystyle F_x = 0 , F_y = 0$

$\displaystyle F_x = k[\frac{q^2}{a^2}+ \frac{q^2}{(a\sqrt2)^2 }.\frac{1}{\sqrt2} +\frac{q Q}{(a/\sqrt2)^2}.\frac{1}{\sqrt2} ] = 0$

$\displaystyle Q = -\frac{q}{4}(1 + 2\sqrt2)$

Similarly , If Fy=0

$\displaystyle Q = -\frac{q}{4}(1 + 2\sqrt2)$