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Home/ Questions/Q 7131
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Ashok0210
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Ashok0210
Asked: February 13, 20212021-02-13T04:26:42+00:00 2021-02-13T04:26:42+00:00In: Class 12 Physics

Two point charges q₁ = 3 μC and q₂ = –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?

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Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Q-1.8
NCERT Solutions for Class 12th Physics

2020-2021chapter 1class 12class 12 physicselectric charges and fieldsncertncert solutions for class 12science
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      Pritika2411
      2021-02-16T11:32:54+00:00Added an answer on February 16, 2021 at 11:32 am

      (a).

      Let us imagine two points A and B such that AO=OB and O is centre of line AB.

      The distance between the two charges  is AB=20cm

      Therefore AO=OB=10cm

      Then net electric field at point O=E (say)

      Electric field at point O as a result of +3µC charge will be =E₁ (say)

      Where (1/4πε0) =9x 10⁹ and  ε0   =Permittivity of free space

      and OA =10cm =10x 10⁻²Nm 2C⁻²

      Then E₁= (1/4πε0) x 3×10⁻⁶/(OA)²

      = (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)²       along OB

      Electric field at point O as a result of -3µC charge will be =E2 (say)

      Then magnitude of  E₂(absolute value)= (1/4πε0) x ( -3×10⁻⁶)/(OB)²

      = (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)²       along OB

      Therefore E= E₁+E2= 2 x (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)²       along OB

      Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.

      Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹

      =5.4 x 10⁶ NC⁻¹ along OB

      Therefore, the electric field at mid-point O is 5.4 x 106 NC⁻¹ along OB.

       

      (b).

      A test charge of amount 1.5 x 10-9 C is placed at mid-point O.

      q= 1.5 x 10⁻9C

      Force experienced by the test charge = F (say)

      Then F = qE

      = 1.5 x 10⁻9 x 5.4 x 106 = 8.1 x 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 x 10-3 N along OA.

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