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Physics Question on Electric Charges and Coulomb's Law

Two point charges qA=3µCq_A = 3µC and qB=−3µCq_B = −3µC 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 1.5×10−9C1.5 × 10^{−9} C is placed at this point, what is the force experienced by the test charge?

Answer

(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+3µC charge,
E1=14πε0.3×10−6(OA)2 = 14πε0.3×10−6(10×10−2)2NC−1  along OBE_1 = \frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OA)^2}\, =\, \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1} \,\,along\,OB
Where, ε0ε_0 = Permittivity of free space and \frac{1}{ 4πε_0}$$= 9 × 10^9 Nm^2C^{-2}
Therefore,
Magnitude of electric eld at point O caused by −3µC−3µC charge,
E2=∣14πε0.3×10−6(OB)2∣=14πε0.3×10−6(10×10−2)2NC−1  along OBE_2 = |\frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OB)^2} |= \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1} \,\,along\,OB
E=E1+E2=2×14πε0.3×10−6(OB)2 along OBE = E_1 + E_2 = 2 × \frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OB)^2} \, along\,OB
[ Since t e magnitudes of E1E_1 and E2E_2 are equal and in t e same direction]
E=2×9×109×14πε0.3×10−6(10×10−2)2NC−1E = 2 × 9 × 10^9 × \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1}
=5.4×106NC−1  along  OB= 5.4 × 10^6 NC^{-1} \,\,along\,\,OB
Therefore, the electric field at mid-point O is 5.4×106NC−15.4 × 10^6 NC^{−1} along OB

(b) A test charge of amount 1.5×10−9C1.5 × 10^{−9} C is placed at mid – point O.
q=1.5×10−9Cq = 1.5 × 10^{−9} C
Force experienced by the test charge = FF
F=qEF = qE
=1.5×10−9×5.4×106= 1.5 × 10^{−9} × 5.4 × 10^ 6
=8.1×10−3N= 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 8.1×10−3N8.1 × 10^{−3} N along OA.