Question

2 point charges of +16 microcoulomb and -9 microcoulomb are placed 8 cm apart in air. Determine the position of the point where resultant electric field is 0 ( zero).justify with diagram.

Answer 1

The position of the point where the resultant electric field is 0 is 24 cm from the negative charge and 32 cm from the positive charge.

Explanation:

As the both the charges have opposite sign that means they are acting in opposite directions, so the zero electric field point is not possible between the two charges , it should occur in any one side. Let us consider the point P where the electric field is zero at a distance x from the negative charge.

So, the diagram is  given below.

Let the point A has the positive charge +16 μC and the point B has the negative charge -9 μC. The distance between the points A and B is 8 cm. Let us consider a point P at a distance x from B. P is the point where the electric field is zero.

The electric field acting on a point due to charge q is,

$E=k \times \frac{q}{(r)^{2}} \rightarrow(1)$

Here k is the constant of proportionality, q is the charge and r is the distance of the point from the charge.

Thus, the electric field acting on point P due to the positive charge at A is,

$E_{1}=k \times \frac{16}{(8+x)^{2}} \rightarrow(2)$

Similarly, the electric field acting on point P due to the negative charge at B is,

$E_{2}=-k \times \frac{9}{(x)^{2}} \rightarrow(3)$

The resultant electric field should be zero at P, so,

$|E|=E_{1}+E_{2}=0$

$\left(k \times \frac{16}{(8+x)^{2}}\right)-\left(k \times \frac{9}{(x)^{2}}\right)=0$

$k \times\left[\frac{16}{(8+x)^{2}}-\frac{9}{(x)^{2}}\right]=0$

$\frac{16}{(8+x)^{2}}-\frac{9}{(x)^{2}}=0$

$\frac{16}{(8+x)^{2}}=\frac{9}{(x)^{2}}$

Taking square root on both the sides,

$\frac{4}{8+x}=\frac{3}{x}$

$4x=3(8+x)$

$4x=24+3x$

$4x-3x=24$

$x=24 \ cm$

Thus, the point P should be at a distance of 24 cm from the negative charge and (24+8) cm distance from the positive charge.

So the distance between PA is 32 cm and PB is 24 cm.

Answer 2

Explanation:

The position of the point where the resultant electric field is 0 is 24 cm from the negative charge and 32 cm from the positive charge.

Explanation:

As the both the charges have opposite sign that means they are acting in opposite directions, so the zero electric field point is not possible between the two charges , it should occur in any one side. Let us consider the point P where the electric field is zero at a distance x from the negative charge.

So, the diagram is given below.

Let the point A has the positive charge +16 μC and the point B has the negative charge -9 μC. The distance between the points A and B is 8 cm. Let us consider a point P at a distance x from B. P is the point where the electric field is zero.

The electric field acting on a point due to charge q is,

E=k \times \frac{q}{(r)^{2}} \rightarrow(1)E=k×

(r)

2

q

→(1)

Here k is the constant of proportionality, q is the charge and r is the distance of the point from the charge.

Thus, the electric field acting on point P due to the positive charge at A is,

E_{1}=k \times \frac{16}{(8+x)^{2}} \rightarrow(2)E

1

=k×

(8+x)

2

16

→(2)

Similarly, the electric field acting on point P due to the negative charge at B is,

E_{2}=-k \times \frac{9}{(x)^{2}} \rightarrow(3)E

2

=−k×

(x)

2

9

→(3)

The resultant electric field should be zero at P, so,

|E|=E_{1}+E_{2}=0∣E∣=E

1

+E

2

=0

\left(k \times \frac{16}{(8+x)^{2}}\right)-\left(k \times \frac{9}{(x)^{2}}\right)=0(k×

(8+x)

2

16

)−(k×

(x)

2

9

)=0

k \times\left[\frac{16}{(8+x)^{2}}-\frac{9}{(x)^{2}}\right]=0k×[

(8+x)

2

16

−

(x)

2

9

]=0

\frac{16}{(8+x)^{2}}-\frac{9}{(x)^{2}}=0

(8+x)

2

16

−

(x)

2

9

=0

\frac{16}{(8+x)^{2}}=\frac{9}{(x)^{2}}

(8+x)

2

16

=

(x)

2

9

Taking square root on both the sides,

\frac{4}{8+x}=\frac{3}{x}

8+x

4

=

x

3

4x=3(8+x)4x=3(8+x)

4x=24+3x4x=24+3x

4x-3x=244x−3x=24

x=24 \ cmx=24 cm

Thus, the point P should be at a distance of 24 cm from the negative charge and (24+8) cm distance from the positive charge.

So the distance between PA is 32 cm and PB is 24 cm.