Question

two charges of +10microcoulomb and +40microcoulomb respectively are placed 12cm apart find the position of the point where electric field is zero.

Answer 1

The field will be zero at a distance of 4 cm from the charge of 10micro c

Answer 2

Answer:  

The point at which electric field is null is at 4 cm away from A and 8 cm away from B.

Solution:

As both positive charges are separated by a distance of 12 cm apart, the electric field will be acting opposite to each other only. So let us consider a point C placed between two points A and B which have charges 10 micro coulomb and 40 micro coulomb.

The electric field acting at a point due to charge q is

$E=\frac{k q}{d^{2}}$

Here k is the proportionality constant =9 \times 10^{9} \mathrm{N} and q is the charge and d is the distance between the charge and the point.

So the electric field acting on C due to 10 μC is  

$E=\frac{10 \times 10^{-6} k}{(A C)^{2}}$

Similarly, the electric field acting on C due to 40 μC is  

$E=\frac{40 \times 10^{-6} k}{(12-A C)^{2}}$

As, 12 is the length of AB.

On equation both equations, we can determine the position of C. So,

$\frac{10 \times 10^{-6} k}{(A C)^{2}}=\frac{40 \times 10^{-6} k}{(12-A C)^{2}}$

$(12-A C)^{2}=\frac{40}{10} \times(A C)^{2}$

$(12-A C)^{2}=4(A C)^{2}$

Taking square root on both sides, we get

$12-A C=2 A C$

$12=3 A C$

$A C=\frac{12}{3}$

$A C=4$

$B C=12-4=8$

Thus the point at which electric field is null is at 4 cm away from A and 8 cm away from B.