Questions: La energla potencial del Eistema mostrado en la figura, si se anade una carga electrica a localizada en el centro del cuadrado de la figura 1 (punto ol es: (A) [ fracsqrt2 g q^prime4 pi epsilon0^a ] (B) [ fracq q^circ4 pi epsilon0^a ] (C) [ frac2 q q^prime4 pi epsilon0^a ] (D) 0 J (E) [ -fracsqrt2 q q^circ4 pi epsilon0 a^a ]

$La energla potencial del Eistema mostrado en la figura, si se anade una carga electrica a localizada en el centro del cuadrado de la figura 1 (punto ol es: (A) [ fracsqrt2 g q^prime4 pi epsilon0^a ] (B) [ fracq q^circ4 pi epsilon0^a ] (C) [ frac2 q q^prime4 pi epsilon0^a ] (D) 0 J (E) [ -fracsqrt2 q q^circ4 pi epsilon0 a^a ]$

Transcript text: La energla potencial del Eistema mostrado en la figura, si se anade una carga electrica a localizada en el centro del cuadrado de la figura 1 (punto ol es: (A) \[ \frac{\sqrt{2} g q^{\prime}}{4 \pi \epsilon_{a}^{a}} \] (B) \[ \frac{q q^{\circ}}{4 \pi t_{0}{ }^{a}} \] (c) \[ \frac{2 q q^{\prime}}{4 \pi \theta_{0}^{a}} \] (D) 0 J (c) \[ -\frac{\sqrt{2} q q^{\circ}}{4 \pi t a^{a}} \]

Solution

Solution Steps

Step 1: Identify the charges and their positions

The charges are located at the corners of a square with side length $ a $. The charges are:

$ +q $ at (a, a)
$ +q $ at (a, -a)
$ -2q $ at (-a, a)
$ -2q $ at (-a, -a)

Step 2: Determine the distance from the center to each charge

The distance from the center of the square (point O) to any corner is given by the formula for the diagonal of a square divided by 2: \[ r = \frac{\sqrt{2}a}{2} = \frac{a\sqrt{2}}{2} \]

Step 3: Calculate the potential at the center due to each charge

The electric potential $ V $ at a point due to a charge $ Q $ is given by: \[ V = \frac{kQ}{r} \] where $ k = \frac{1}{4\pi\epsilon_0} $.

For each charge at the center:

Potential due to $ +q $ at (a, a) and (a, -a): \[ V_1 = \frac{kq}{\frac{a\sqrt{2}}{2}} = \frac{2kq}{a\sqrt{2}} = \frac{2kq\sqrt{2}}{2a} = \frac{kq\sqrt{2}}{a} \]
Potential due to $ -2q $ at (-a, a) and (-a, -a): \[ V_2 = \frac{k(-2q)}{\frac{a\sqrt{2}}{2}} = \frac{-2kq}{\frac{a\sqrt{2}}{2}} = \frac{-4kq}{a\sqrt{2}} = \frac{-2kq\sqrt{2}}{a} \]

Step 4: Sum the potentials at the center

The total potential at the center is the sum of the potentials due to each charge: \[ V_{\text{total}} = 2 \left( \frac{kq\sqrt{2}}{a} \right) + 2 \left( \frac{-2kq\sqrt{2}}{a} \right) \] \[ V_{\text{total}} = \frac{2kq\sqrt{2}}{a} - \frac{4kq\sqrt{2}}{a} \] \[ V_{\text{total}} = \frac{-2kq\sqrt{2}}{a} \]

Step 5: Calculate the potential energy of the system

The potential energy $ U $ of a charge $ q $ in a potential $ V $ is given by: \[ U = qV \] For the charge $ q $ placed at the center: \[ U = q \left( \frac{-2kq\sqrt{2}}{a} \right) \] \[ U = \frac{-2kq^2\sqrt{2}}{a} \]