ચાર -Q વિદ્યુતભારોને ચોરસના ચાર ખૂણાઓ આગળ મૂક | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Step 1: Calculate all the required distances between the charges :

Let the side length of the square be $a$.

Refer Figure, In $	riangle A C D$,

Vedclass · Accepted Answer

$\frac{Q}{4}\,(1\,\, + \,\,2\,\sqrt 2 )$

Vedclass · Answer

Step 1: Calculate all the required distances between the charges :

Let the side length of the square be $a$.

Refer Figure, In $	riangle A C D$,

$\Rightarrow AD ^2+ CD ^2= AC ^2$

$\Rightarrow a ^2+ a ^2= AC ^2 \quad \Rightarrow AC =\sqrt{2} a$

However, $O C=\frac{ AC }{2}=\frac{ a }{\sqrt{2}}$

Step $2$ : Equilibrium of central charge

By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.

Hence, Net force on central charge will always be zero, irrespective of value of charge q.

Therefore, to find value of $q$ we have to check equilibrium of any one charge at the corner.

Step 3 : Force due to all the charges at point $C$

Force on charge at $C$ due to $B, \quad F_1=\frac{K^2}{a^2}$

Force on charge at $C$ due to $D, \quad F_2=\frac{K^2}{a^2}$

Force on charge at $C$ due to $A , \quad F _4=\frac{ KQ ^2}{ AC ^2}=\frac{ KQ ^2}{2 a ^2}$

Force on charge at $C$ due to $q$ at centre, $F_3=\frac{K q Q}{O C^2}=\frac{2 K q Q}{a^2}$

Step $4$ : Apply the equilibrium condition at $C$ :

For the system to be in equilibrium, net force acting on charge at $C$ must be zero.

So, $\vec{F}_3+\vec{F}_1+\vec{F}_2+\vec{F}_4=0$

Resultant of $F_1$ and $F_2$ (Along OC, by symmetry) $=\sqrt{F_1^2+F_2^2}=\sqrt{2} F_1$

Since $\left(\left| F _1ight|=\left| F _2ight|ight)$

Also, $F _3 \& F _4$ are along OC, Therefore, magnitudes of sum of these forces should be zero.

$\Rightarrow \quad\left| F _3ight|+\sqrt{2}\left| F _1ight|+\left| F _4ight|=0$

$\Rightarrow \frac{2 K q Q}{a^2}+\sqrt{2} \frac{K Q^2}{a^2}+\frac{K Q^2}{2 a^2}=0$

$\Rightarrow 2 q =-\left(\sqrt{2} Q +\frac{ Q }{2}ight)$

$\Rightarrow q =\frac{ Q }{4}(1+2 \sqrt{2})$

Similar Questions

વિદ્યુતક્ષેત્ર રેખાઓ અને સ્થિતિમાન પૃષ્ઠ વચ્ચેનો ખૂણો .........$^o$ છે.