Respuesta :
Answer with Explanation:
We are given that
[tex]q_1=q_2=q_3=q_4=q=9\mu C=9\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
Side of square,d=4 m
Distance of charge from the center of square,d=[tex]\frac{1}{2}\sqrt{4^2+4^2}=2\sqrt 2m[/tex]
Potential is a scalar quantity therefore, we consider sign .
a.When all charges are positive
Then, Potential at center of square,V=[tex]4\times \frac{kq}{d}[/tex]
Where [tex]k=9\times 10^9[/tex]
[tex]V=4\times \frac{9\times 10^9\times 9\times 10^{-6}}{2\sqrt 2}=1.15\times 10^5 V[/tex]
b.Potential at center of square,V=[tex]3\frac{kq}{d}-\frac{kq}{d}=\frac{2kq}{d}[/tex]
[tex]V=\frac{2\times 9\times 10^9\times 9\times 10^{-6}}{2\sqrt 2}=5.7\times 10^{4} V[/tex]
c.Potential at center of square,V=[tex]\frac{2kq}{d}-\frac{2kq}{d}=0[/tex]
Answer:
(a) 81000 V, (b) 40500 V, (c) 0 V
Explanation:
Given:
side of square, a = 4 m
charge on each corner , q = + 9 micro coulomb
The formula for the potential is
[tex]V = \frac{K q}{r}[/tex]
where, k is the Coulombic constant.
(a)
All the charges are positive:
Let V1, V2, V3 and V4 be the potential at the centre O due to the charges at 1, 2,3 and 4.
[tex]V_{1} = \frac{K q}{a}[/tex]
[tex]V_{2} = \frac{K q}{a}[/tex]
[tex]V_{3} = \frac{K q}{a}[/tex]
[tex]V_{4} = \frac{K q}{a}[/tex]
Total potential at the centre of square is
[tex]V = V_{1}+ V_{2}+ V_{3}+ V_{4}[/tex]
[tex]V = \frac{4K q}{a}[/tex]
[tex]V = \frac{9\times 10^{9}\times 4\times 9\times 10^{-6}}{4}[/tex]
V = 81000 V
(b)
Three charge are positive and one is negative
[tex]V_{1} = \frac{K q}{a}[/tex]
[tex]V_{2} = \frac{K q}{a}[/tex]
[tex]V_{3} = \frac{K q}{a}[/tex]
[tex]V_{4} = -\frac{K q}{a}[/tex]
Total potential at the centre of the square is
[tex]V = V_{1}+ V_{2}+ V_{3}+ V_{4}[/tex]
[tex]V = \frac{2K q}{a}[/tex]
[tex]V = \frac{9\times 10^{9}\times 2\times 9\times 10^{-6}}{4}[/tex]
V = 40500 V
(c)
two charge are positive and the other two are negative
[tex]V_{1} = \frac{K q}{a}[/tex]
[tex]V_{2} = \frac{K q}{a}[/tex]
[tex]V_{3} = -\frac{K q}{a}[/tex]
[tex]V_{4} = -\frac{K q}{a}[/tex]
Total potential at the centre of the square is
[tex]V = V_{1}+ V_{2}+ V_{3}+ V_{4}[/tex]
V = 0 V
