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
]
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=22a=2a2
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=rkQ
where k=4πϵ01.
For each charge at the center:
Potential due to +q at (a, a) and (a, -a):
V1=2a2kq=a22kq=2a2kq2=akq2
Potential due to −2q at (-a, a) and (-a, -a):
V2=2a2k(−2q)=2a2−2kq=a2−4kq=a−2kq2
Step 4: Sum the potentials at the center
The total potential at the center is the sum of the potentials due to each charge:
Vtotal=2(akq2)+2(a−2kq2)Vtotal=a2kq2−a4kq2Vtotal=a−2kq2
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(a−2kq2)U=a−2kq22