Two positive charges of magnitude q are place | vedclass

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Question

(A) Let the side length of the square be $2a$. The charges are placed at the corners. Let the side with positive charges be along the $x$-axis at $y=2

Vedclass · Accepted Answer

$\frac{1}{4\pi \varepsilon_{0}} \frac{2qQ}{a} \left( 1 - \frac{1}{\sqrt{5}} \right)$

Vedclass · Answer

(A) Let the side length of the square be $2a$. The charges are placed at the corners. Let the side with positive charges be along the $x$-axis at $y=2a$. The midpoint of this side is at $(a, 2a)$. The centre of the square is at $(a, a)$.Initial potential at the midpoint $i(a, 2a)$:The distances to the two positive charges are $a$ and $a$. The distances to the two negative charges are $\sqrt{a^2 + (2a)^2} = a\sqrt{5}$ and $a\sqrt{5}$.$V_i = \frac{kq}{a} + \frac{kq}{a} - \frac{kq}{a\sqrt{5}} - \frac{kq}{a\sqrt{5}} = \frac{2kq}{a} \left( 1 - \frac{1}{\sqrt{5}} \right)$.Final potential at the centre $f(a, a)$:The distance to each of the four corners is $\sqrt{a^2 + a^2} = a\sqrt{2}$.$V_f = \frac{kq}{a\sqrt{2}} + \frac{kq}{a\sqrt{2}} - \frac{kq}{a\sqrt{2}} - \frac{kq}{a\sqrt{2}} = 0$.By the work-energy theorem,the change in kinetic energy $\Delta K = W_{ext} = -W_{electric} = -Q(V_f - V_i) = Q(V_i - V_f)$.Since the charge starts from rest,$K_i = 0$,so $K_f = Q(V_i - V_f) = Q \left[ \frac{2kq}{a} \left( 1 - \frac{1}{\sqrt{5}} \right) - 0 \right] = \frac{1}{4\pi \varepsilon_{0}} \frac{2qQ}{a} \left( 1 - \frac{1}{\sqrt{5}} \right)$.

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