Question 8:
Two point charges qA= 3 µC and qB = −3 µC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
Solution:
Two point charges qA= 3 µC and qB = −3 µC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
qA = 3 µC = 3 × 10 – 6 C
qB = −3 µC = - 3 × 10 – 6 C
Distance AB = 20 cm = 0.2 m
In this problem, O is the mid-point of line AB.
OB = OA = 0.2/2 = 0.1 m
The electric field produced by the charge Q at a point r is given as
Plug the values we get
Electric field at point O caused by +3µC charge,
Magnitude of electric field at point O caused by −3µC charge,
The direction of electric field is always positive to negative so that
Direction of electric field will in direction of OB
Total electric field due to both charge
E total = E1+ E2
Plug 1/4πε0 = 9 × 109 Nm2C-2
We get
E total = 5.4 × 106 N/C along OB
Answer
The electric field at mid-point O is 5.4 × 106 N C−1 along the direction OB.
Given that
(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
q = 1.5 × 10−9C
Force experienced by the test charge = F
Use formula of electrostatic force of attraction
F = qE
Plug the values in this formula we get
F = 1.5 × 10−9× 5.4 × 106 = 8.1 × 10−3 N
Here placed charge is negative so negative charge at point B will repel while positive change at point A will attract. Hence the direction of force will along OA
Answer
The force experienced by the test charge = 8.1 × 10−3 N along the direction of OA.
Wtemptipulcji_1982 Max Mullen https://wakelet.com/wake/EDpkLrw2TPWQat4Zl90Az
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