The figure below shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm. (Let q1 = +1.00 nC, q2 = +8.00 nC, q3 = +5.00 nC, and 94-+2.00 nC. Assume q3 is located at the origin and x axis is to the right and the ty axis is up along the page. Express your answers in vector form.) 91 42 73 (a) What is the electric field at the location of the sphere with charge +5.00 nC? N/C (b) What is the total electric force exerted on the sphere with charge +5.00 nC by the other three spheres?
Solution
Assume q 3 is located at the origin O(0,0),then the other co-ordinate is q 1 = 1nC =1*10 -9 C at R(0,d),q 2 =8 nC =8*10 -9 C at Q(d,d) and q 4 =2 nC =2*10 -9 C respectively.Then,
All charges are positive in nature, so they will be repel to each other.
According to Coulomb\'s law, the forces experienced by a charge q3 =+5*10 -9 at point O due other charges are given as
F P =(1/4?€)[5*10 -9 *2*10 -9 /(.25) 2 ]=0.144*10 -13
F R =(1/4?€)[5*10 -9 *1*10 -9 /(.25) 2 ]=0.072*10 -13
F Q =(1/4?€)[5*10 -9 *8*10 -9 /(.25?2) 2 ] =0.288*10 -13
F P and F R are perpendicular.So,
Resultant , F RES =?[(F P ) 2 +(F R ) 2 ]
=?(0.144*10 -13 ) 2 +(0.72*10 -13 ) 2
= ?[0.0207+0.5184]*(10 -13 ) 2
=?5391*10 -13
= .76*10 -13
Now, F R and F OQ are in the same direction.
Resultant F Resultant = .76*10 -13 +0.288*10 -13
=1.048*10 -13 N
E letric Field, E =F/Q=1.048*10 -13 /5*10 -9
=(1.048/5)*10 (-13+9 )
= .2096*10 -4 N/C
.
The figure below shows four small charged spheres arranged at the corn.docx
1. The figure below shows four small charged spheres arranged at the corners of a square with side
d = 25.0 cm. (Let q1 = +1.00 nC, q2 = +8.00 nC, q3 = +5.00 nC, and 94-+2.00 nC. Assume q3 is
located at the origin and x axis is to the right and the ty axis is up along the page. Express your
answers in vector form.) 91 42 73 (a) What is the electric field at the location of the sphere with
charge +5.00 nC? N/C (b) What is the total electric force exerted on the sphere with charge
+5.00 nC by the other three spheres?
Solution
Assume q 3 is located at the origin O(0,0),then the other co-ordinate is q 1 = 1nC =1*10 -9
C at
R(0,d),q 2 =8 nC =8*10 -9
C at Q(d,d) and q 4 =2 nC =2*10 -9
C respectively.Then,
All charges are positive in nature, so they will be repel to each other.
According to Coulomb's law, the forces experienced by a charge q3 =+5*10 -9
at point O due
other charges are given as
F P =(1/4?€)[5*10 -9
*2*10 -9
/(.25) 2
]=0.144*10 -13
F R =(1/4?€)[5*10 -9
*1*10 -9
/(.25) 2
]=0.072*10 -13
F Q =(1/4?€)[5*10 -9
*8*10 -9
/(.25?2) 2
] =0.288*10 -13
F P and F R are perpendicular.So,
Resultant , F RES =?[(F P ) 2
+(F R ) 2
]
=?(0.144*10 -13
) 2
+(0.72*10 -13
) 2
=
?[0.0207+0.5184]*(10 -13
) 2
=?5391*10 -13
=
.76*10 -13
Now,
F R and F OQ are in the same direction.
2. Resultant F Resultant = .76*10 -13
+0.288*10 -13
=1.048*10 -13
N
E letric Field, E =F/Q=1.048*10 -13
/5*10 -9
=(1.048/5)*10 (-13+9 )
=
.2096*10 -4
N/C
![The figure below shows four small charged spheres arranged at the corners of a square with side
d = 25.0 cm. (Let q1 = +1.00 nC, q2 = +8.00 nC, q3 = +5.00 nC, and 94-+2.00 nC. Assume q3 is
located at the origin and x axis is to the right and the ty axis is up along the page. Express your
answers in vector form.) 91 42 73 (a) What is the electric field at the location of the sphere with
charge +5.00 nC? N/C (b) What is the total electric force exerted on the sphere with charge
+5.00 nC by the other three spheres?
Solution
Assume q 3 is located at the origin O(0,0),then the other co-ordinate is q 1 = 1nC =1*10 -9
C at
R(0,d),q 2 =8 nC =8*10 -9
C at Q(d,d) and q 4 =2 nC =2*10 -9
C respectively.Then,
All charges are positive in nature, so they will be repel to each other.
According to Coulomb's law, the forces experienced by a charge q3 =+5*10 -9
at point O due
other charges are given as
F P =(1/4?€)[5*10 -9
*2*10 -9
/(.25) 2
]=0.144*10 -13
F R =(1/4?€)[5*10 -9
*1*10 -9
/(.25) 2
]=0.072*10 -13
F Q =(1/4?€)[5*10 -9
*8*10 -9
/(.25?2) 2
] =0.288*10 -13
F P and F R are perpendicular.So,
Resultant , F RES =?[(F P ) 2
+(F R ) 2
]
=?(0.144*10 -13
) 2
+(0.72*10 -13
) 2
=
?[0.0207+0.5184]*(10 -13
) 2
=?5391*10 -13
=
.76*10 -13
Now,
F R and F OQ are in the same direction.](https://image.slidesharecdn.com/thefigurebelowshowsfoursmallchargedspheresarrangedatthecorn-230209233759-11ae0b45/85/The-figure-below-shows-four-small-charged-spheres-arranged-at-the-corn-docx-1-320.jpg)
