NCERT Solutions
Class 12 Physics
Electric Charges and Fields
Q8.
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?
(a) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB = 20 cm
∴ AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
E1 = (1/4πε0). ((3 × 10−6)/ (OA) 2)
= (1/4πε0). (3 × 10−6)/ ((10 × 10−2)2) NC−1 along OB
Where, ε0 = Permittivity of free space and (1/4πε0) = 9 × 109 Nm2C−2
Therefore,
Magnitude of electric field at point O caused by −3μC charge,
E2 = | (1/4πε0) (−3 × 10−6)/ ((OB) 2)|
= (1/4πε0) ((3 × 10−6)/ (10 × 10−2)2) NC−1 along OB
∴ E = E1 + E2
= (2 × (1/4πε0)). (3 × 10−6 (10 × 10−2)2)NC−1 along OB
[Since the magnitudes of E1 and E2 are equal and in the same direction]
∴ E = (2 × 9 × 109) × ((3 × 10−6)/ ((10 × 10−2)2) NC−1
= (5.4 × 106 NC–1 ) along OB
Therefore, the electric field at mid-point O is 5.4 × 106 N C−1along OB.
(b) A test charge of amount 1.5 × 10−9 C is placed at mid – point O.
q = 1.5 × 10−9 C
Force experienced by the test charge = F
∴ F = (q E) = 1.5 × 10−9 × 5.4 × 106 = 8.1 × 10−3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.