Question

Two point charges +9 μc and –6 μc are placed 16 cm apart in air. Find the position           

       of the point at which the resultant electric field is zero.​

Answer 1

Given ,

The two point charges +9 μc and –6 μc are located at A and B and separated by 16 cm

Let ,

• P be the point where electric field is zero
• The distance b/w A and P be x
• So , the distance b/w P And B will be (16-x)

The electric field at point P will be zero if ,

$\mathtt{ \large\fbox{E_{ q_{1}} = E_{ q_{2}}}}$

Thus ,

$\sf \hookrightarrow \frac{k \times 9 \times {(10)}^{ - 6} }{ {(x)}^{2} } = \frac{k \times 6 \times {(10)}^{ - 6} }{ {(16 - x)}^{2} } \\ \\ \sf \hookrightarrow {(16 - x)}^{2} \times 9 = {(x)}^{2} \times 6 \\ \\\sf \hookrightarrow (16 - x) \times 3 = \sqrt{6}x \\ \\\sf \hookrightarrow 48 - 3x = \sqrt{6}x \\ \\\sf \hookrightarrow 48 = \sqrt{6}x + 3 \\ \\\sf \hookrightarrow 48 = x( \sqrt{6} + 3) \\ \\\sf \hookrightarrow x = \frac{48}{ \sqrt{6} + 3} \: \: \: \: \{ \because \sqrt{6} = 2.4\} \\ \\\sf \hookrightarrow x = \frac{48}{5.4} \\ \\\sf \hookrightarrow x = 8.9 \: \: cm$

Hence , the electric field will be zero at 8.9 cm from the right of first charge

Answer 2

Answer:

8.9cm , 7.1 cm

Explanation:

Let the distance from 9 micro coulomb be X cm

The total distance is 16 cm, so the distance from -6 micro coulomb will be (16-X) cm

We have to find the spot where :

• Eq1=Eq2

According to the formula,

$\frac{k \times 9 \times {10}^{ - 6} }{ {x}^{2} } = \frac{k \times 6 \times {10}^{ - 6} }{(16 - x )^{2} }$

Cancelling k and 10^-6 on both sides,

$\frac{9}{x ^{2} } = \frac{6}{(16 - x) ^{2} }$

By cross multiplying,

$6x ^{2} = 9(256 + x ^{2} - 32x)$

By solving the following equation,

We get X=8.9cm

8.9cm

16-X= 7.1 cm