Two point charges, q1 = 2.1 imes 10⁻⁵ ext C and q2 = 4q1, are separated by a distance of 10 cm. Calculate the electric field intensity at point P, which is 8 cm from q1 and 6 cm from q2. Assume k = 9

The charges are:
$q_1 = 2.1 \times 10^{-5} \text{ C}$
$q_2 = 4q_1 = 4 \times (2.1 \times 10^{-5} \text{ C}) = 8.4 \times 10^{-5} \text{ C}$

The distances are:
Distance between $q_1$ and $q_2$ is $d = 10 \text{ cm} = 0.10 \text{ m}$.
Distance from $q_1$ to the point P is $r_1 = 8 \text{ cm} = 0.08 \text{ m}$.
Distance from $q_2$ to the point P is $r_2 = 6 \text{ cm} = 0.06 \text{ m}$.

The electrostatic constant is $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

We check if the point P forms a right-angled triangle with the charges $q_1$ and $q_2$.
$r_1^2 + r_2^2 = (0.08 \text{ m})^2 + (0.06 \text{ m})^2 = 0.0064 \text{ m}^2 + 0.0036 \text{ m}^2 = 0.0100 \text{ m}^2$.
$d^2 = (0.10 \text{ m})^2 = 0.0100 \text{ m}^2$.
Since $r_1^2 + r_2^2 = d^2$, the point P forms a right-angled triangle with $q_1$ and $q_2$, with the right angle at point P. This means the electric field vectors $E_1$ and $E_2$ at point P will be perpendicular to each other.

The formula for the electric field intensity due to a point charge is $E = \frac{k|q|}{r^2}$.

For charge $q_1$:

For charge $q_2$:

Since $