Question

Two point charges q_A = 3 μC and q_B = −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?

Answer 1

Given:

q_A ​= 3μC, q_B​ = −3μC, r = 20 cm 0.2 m, E_R ​= ?, F = ?

we have E = \(\frac 1{4\pi \epsilon_0} \frac 2{r^2}\)

Electric intensity due to q_A​ on P

(a) Resultant field, E_R​ = E_A​ + E_B​

= 2.7 × 10⁶ + 2.7 × 10⁶

ER ​= 5.4 × 10⁶ N/C along AB

(b) F = E_q ​= 5.4 × 10⁶ × 1.5 × 10⁻⁹

F = 8.1 × 10⁻³N along BA.

Answer 2

(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,

Where, ₀ = Permittivity of free space and 1/4πε_o=9×10⁹ Nm²C⁻²
Therefore,
Magnitude of electric field at point O caused by −3μC charge,

[ Since the magnitudes of E₁ and E₂ are equal and in the same direction  ]

Therefore, the electric field at mid-point O is 5.4 × 10⁶ N C⁻¹ along OB.

(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

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.