Two charges, A (-0.144 nC) and B (0.256 nC), are located at (-16 cm, 0 cm ) and (0 cm,12 cm), respectively. The magnitude of electric field at point (-16 cm,12 cm) due to these two charges is close to:

Concept:

• Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles.
• \(E = \frac{KQ}{r^2}\)
• Where, \(K = \frac{1}{4\pi \epsilon _0}\) = 9 × 10⁹ Nm²C^-2, Q is the charge, r is the distance from the charge.
• According to the incoming charge’s nature, the electric field will either attract or repel the charge.
• The electric field at a point is a vector quantity 
• \(\overrightarrow E= \overrightarrow E_1 + \overrightarrow E_2\)
• Magnitude of \(E =\sqrt{E_1^2 + E_2 ^2}\) when angle between E₁ and E₂ is 90° 
• Distance between two points A (x₁, y₁) and B (x₂, y₂) is equal to \(\sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2} \)

Calculation:

Given charge A = -0144 × 10^-9 C, B = 0.256 × 10^-9 C

Let the point (-16 cm,12 cm) is P,

Distance between points A (-16 cm, 0 cm) and P (-16 cm,12 cm), r_A = \(\sqrt {(-16 - (-16))^2 + (0-12)^2} = 12\) cm

Electric field due to charge A on point P is

\(E_A = \frac{9 \times ​​10^9 \times(-0.144 \times 10^{-9})}{0.12^2}\) = -90

Now, distance between points (0 cm, 12 cm) and P (-16 cm,12 cm), r_B = \(\sqrt {(0 - (-16))^2 + (12-12)^2} = 16\) cm

Electric field due to charge B on point P is

\(E_B = \frac{9 \times ​​10^9 \times0.256 \times 10^{-9}}{0.16^2}\) = 90

Also angle between E_A and E_B is 90° then,

The magnitude of electric field \(E =\sqrt{E_A^2 + E_B ^2}\)

\(\Rightarrow E =\sqrt{(-90)^2 +90^2} \) =127.279 ≈ 127 N/C