Question

Two points charges q₁ = 3μC and q₂ = -3μC 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 charge of magnitude 1.5 × 10⁻⁹ C is placed at the centre, find the force experienced by the test charge.

Answer 1

Electrostatics:

(i). At the mid point of the line AB joining two charges, AO = 10cm or 0.1 m = OB

So, since it is placed at mid point but q₁ and q₂ are opposite, so the electric field will gets add up which will be towards OB ( +ve to - ve ).

$\mathsf{E_{net}}$ = $\mathsf{E_{AO}}$ + $\mathsf{E_{OB}}$

$\mathsf{E_{net}}$ = 2$\mathsf{k*{\dfrac{q}{r^2}}}$ = 2$\mathsf{[9*10^{9}*{\dfrac{3*10^{-6}}{(0.1)^2}}]}$

• $\mathsf{E_{net}}$ = $\mathsf{5.4*10^6 NC^{-1}, along OB}$

(ii). Force experienced by negative test charge, F = qE

$\mathsf{F_{OA} =1.5*10^{-9}* 5.4* 10^6}$

• $\mathsf{F_{OA} = 8.1*10^{-3}N}$

Since, test charge is - ve, so it will be attracted by +ve ( q₁ ) and repelled by - ve ( q₂ ). So, direction of Force will be along OA.

Answer 2

Answer:

the answer is 8.1×10-3 ...