Question

Two charges q₁ = 20 × 10^-6 and q₂ = -20 × 10^-6 C are placed 20 cm apart. Calculate the Electric field intensity at point 20 cm away from both charges.

Answer 1

E₁ = \(\frac1{4\pi\varepsilon_0}\frac{q}{r_1^2}\) 

E₂ = \(\frac1{4\pi \varepsilon_0}\frac{q_2}{r_2^2}\) 

E₁ = 9 x 10⁹ x \(\frac{20\times10^{-6}}{(0.2)^2}\) = 4500 x 10³ ⇒ 4.5 x 10⁶ N/C

E₂ = 9 x 10⁹ x \(\frac{20\times10^{-6}}{(0.2)^2}\) ⇒ 4500 x 10³ ⇒ 4.5 x 10⁶ N/C

Then resultant electric field be

E = E₁ cos \(\theta\) + E₂ cos \(\theta\)

E = 4.5 x 10⁶ cos \(\frac{\pi}3\) + 4.5 x 10⁶ cos \(\frac{\pi}3\) 

E = 4.5 x 10⁶[1/2 + 1/2]

E = 4.5 x 10⁶ N/C