Class-12th Physics, CBSE and UP Board

Electric Charges and Fields,

Chapter-1, Exercise -1, Q-1.8

NCERT Solutions for Class 12th 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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(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 (1/4πε

_{0}) =9x 10⁹ and ε_{0 }=Permittivity of free spaceand OA =10cm =10x 10⁻²Nm

^{ 2}C⁻²Then E₁= (1/4πε

_{0}) x 3×10⁻⁶/(OA)²= (1/4πε

_{0}) x 3×10⁻⁶/(10x 10⁻²)² along OBElectric field at point O as a result of -3µC charge will be =E2 (say)

Then magnitude of E₂(absolute value)= (1/4πε

_{0}) x ( -3×10⁻⁶)/(OB)²= (1/4πε

_{0}) x 3×10⁻⁶/(10x 10⁻²)² along OBTherefore E= E₁+E2= 2 x (1/4πε

_{0}) x 3×10⁻⁶/(10x 10⁻²)² along OBSince 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 10

^{6}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⁻

^{9}CForce experienced by the test charge = F (say)

Then F = qE

= 1.5 x 10⁻

^{9}x 5.4 x 10^{6 }= 8.1 x 10^{-3}NThe 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.