Question

Two charges, Q1=2.3x10-9C and Q2 =3.5x10-9C, are 55mm apart. What is the electric field halfway between them?

Answer 1

Given:

• Q₁ = 2.3 x 10⁻⁹ C
• Q₂ = 3.5 x 10⁻⁹ C
• r = 55 mm = 55 x 10⁻³m
• k ≈ 9 x 10⁹ (approx value has been taken for easy calculations)

Solution:

The electric field produced by two charges placed at distance r apart from each other is given by,

$\to\boxed{\red{\sf{E=\dfrac{kQ_1Q_2}{r^2}}}}$

Now let's substitute the given values in the above equation,

$\to{\sf{E=\dfrac{kQ_1Q_2}{(r/2)^2}}$

$\to{\sf{E=\dfrac{9 \times 10^9 \times 2.3 \times 10^{-9} \times 3.5 \times 10^{-9}}{(55 \times 10^{-3}/2)^2}}$

$\to{\sf{E=\dfrac{72.45 \times 10^{-9}}{756.25 \times 10^{-3}}}$

$\to{\sf{E=95.8 \times 10^{-9} N /m^2 }$

The electric field halfway between Q₁ and Q₂ is 9.58 x 10⁻⁹ Nm².