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NEET-XII-Physics

01: Electric Charges And Fields

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  • #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?
    Ans : null (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 = along OB

    Where,

    = Permittivity of free space



    Magnitude of electric field at point O caused by -3μC charge,

    E2 = = along OB



    = 5.4 × 106 N/C along OB

    Therefore, the electric field at mid-point O is 5.4 × 106 N C-1 along 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 = qE

    = 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.
  • #8-a
    What is the electric field at the midpoint O of the line AB joining the two charges?
    Ans : 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 = along OB

    Where,

    = Permittivity of free space



    Magnitude of electric field at point O caused by -3μC charge,

    E2 = = along OB



    = 5.4 × 106 N/C along OB

    Therefore, the electric field at mid-point O is 5.4 × 106 N C-1 along OB.
  • #8-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?
    Ans : 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 = qE

    = 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.