Two point charges of 25μC and -36μC of charges are separated by a distance of 4m in air. Determine the position of the point at which the
resultant electric field is zero and find the magnitude and sign of charge to be placed at that point so that the system will be in equilibrium? `(20m 500µC)
Answers
Explanation:
Let the two charges placed at two points of line AB as A and B whose distance is 20 cm and O is the mid point of the line AB
Distance between the two charges , AB=20 cm
∴AO=OB
Net elctric at O=E
Electric field at point O caused by +5\mu C charge,
E
1
=
4πε
0
(AO)
2
5×10
−6
=
4πε
0
(10×10
−2
)
2
5×10
−6
N/C toward +5μC
where,
ε
0
=Permittivity of free space
4πε
1
=9×10
9
Nm
2
C
−2
Magnitude of electric field at point O cause by +10\mu C charge,
E
1
=
4πε
0
(AO)
2
10×10
−6
=
4πε
0
(10×10
−2
)
2
10×10
−6
N/C towards +10μC
∴E=E
1
+E
2
=
(10×10
−2
)
2
9×10
9
×15×10
−6
=13.5×10
6
N/C towards +5μC
Chennai ah!!!!!!!!
Explanation:
anna neenga Dhruv Vikram ah paathu irrukeengala??
naa avaroda periya fan...
who is your favorite hero Anna??
Dhruv Vikram na avlo pidikum... chocolate boyyy!!!!!!
and Vikram sir ah um romba romba romba romba romba pidikum ❤️