Question

Two point charges \(q_A = 3µC\) and \(q_B = −3µC\) 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 \(1.5 × 10^{−9} C\) is placed at this point, what is the force experienced by the test charge?

Answer 1

(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,
\(E_1 = \frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OA)^2}\, =\, \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1} \,\,along\,OB\) 
Where, \(ε_0\) = 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 \(−3µC\) charge,
\(E_2 = |\frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OB)^2} |= \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1}  \,\,along\,OB\)
\(E = E_1 + E_2 = 2 ×  \frac{1}{ 4πε_0}.\frac {3 × 10^{-6} }{ (OB)^2} \, along\,OB\)
[ Since t e magnitudes of \(E_1\) and \(E_2\) are equal and in t e same direction]
\(E = 2 × 9 × 10^9 ×  \frac{1 }{ 4πε_0} .\frac{ 3 × 10^{-6}}{(10 × 10^{-2})^2 }NC^{-1}\)
\(= 5.4 × 10^6 NC^{-1} \,\,along\,\,OB\)
Therefore, the electric field at mid-point O is \(5.4 × 10^6 NC^{−1}\) along OB

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(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 = \(F\)
\(F = qE\)
\(= 1.5 × 10^{−9} × 5.4 × 10^ 6\)
\(= 8.1 × 10^{−3} N\)
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.