Question

Two charges Q₁ = +3nC and Q₂ = -4nC are separated by a distance of 50 cm. What is the electric field strength at a point that is 20 cm from Q₁ and 30 cm from Q₂? The point lies between Q₁ and Q₂.

Answer 1

The modulus of electric field of the charge q is $E(r) = k\dfrac{q}{r^2}.$ If we take the direction of field into account, then the fields in question are directed towards the negative charge. Therefore, the modulus of total field will be

$E = E_1+E-2 = k\left( \dfrac{|Q_1|}{r_1^2}+\dfrac{|Q_2|}{r_2^2}\right) = 9\cdot10^9\left( \dfrac{3\cdot10^{-9}}{0.2^2} + \dfrac{4\cdot10^{-9}}{0.3^2}\right) = 1075\,\mathrm{N/C}.$