Two point charges q_A = +2 μC and q_B = –3 p. C are located 20 cm apart in vacuum.
(i) Find the electric field at the midpoint of the line AB joining the two charges, (ii) If a negative test charge of magnitude 1.5 x 10^–9 C is placed at the centre, find the force experienced by the test charge.

Solution

Let, the midpoint of the line joining the two charges be O.

a) Electric field at O due to charge at A-E_{A= $\frac{1}{4 {πε}_{\circ}} . \frac{q_{A}}{r^{2}} = \frac{9 \times 10^{9} \times 2 \times 10^{- 6}}{0 . 1^{2}} N C^{- 1} = 1800 \times 10^{3} N C^{- 1}$}Electric field at O due to charge at B-E_{B= $\frac{1}{4 {πε}_{\circ}} . \frac{q_{B}}{r^{2}} = \frac{9 \times 10^{9} \times - 3 \times 10^{- 12}}{0 . 1^{2}} N C^{- 1} = 2700 \times 10^{- 3} {NC}^{- 1}$}Resultant intensity -E =E_A+ E_B
= $4.5 \times 10^{3} {NC}^{- 1}$

b) If a test charge of magnitude $1.5 \times 10^{- 9} C$ is placed at the centre, then force experienced is $q_{\circ} E$

$1.5 \times 10^{- 9} \times 4.5 \times 10^{3} = 6.75 \times 10^{- 6} C$

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