Question

Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.

Answer 1

Answer:

The electric field intensity is 30000 N/C.

Explanation:

Given:

Magnitude of the point charge is, $q=3\times 10^{-9}\ C$

Distance of the given point from the point charge is, $d=3\ cm=0.03\ m$

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

$E=\frac{kq}{d^2}$

Plug in $k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m$. Solve for 'E'.

$E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C$

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.