Question

calculate the electric field intensity at a point distance 4 meter from a charge of 16 micro coulomb.​

Answer 1

Answer:

We have to search for the null point where the resultant field is zero. Since the two point charges are of similar nature (both positive), there cannot be a null point on the line lying either left of point A or right of point B as in these regions the individual fields produced by the charges are parallel and hence cannot nullify each other. However there is a possibility of null point between the points A and B. Let P be that point (lying at a distance x from point A).

At point P,

Therefore, the resultant electrostatic field is zero at a point (on the line joining the two charges) lying 2m right of the charge 16 μC or 1m left of the charge 4μC

Answer 2

Answer:

The electric field intensity at a distance of  4m from a charge  $q_{1}=16*10^{-6}C$ is $9*10^{3} N/C$

Explanation:

When two charges $q_{2} ,q_{2}$ are separated by a distance $r$ the force attraction or repulsion is given by the formula

$F=K\frac{q_{1} q_{2} }{r^{2} }$

where the value of $K=9*10^{9}$

the electric field is the force per unit charge, that is $q_{2}=1C$

from the question, it is given that the charge is $q_{1}=16*10^{-6}C$ and the distance where the electric field to be calculated is $d=4m$

now the electric field is $E=K\frac{q_{1} }{r^{2} }=\frac{9*10^{9}*16*10^{-6} }{4^{2} }=9*10^{3}N/C$