Question

Two point charges q₁ = 4μC and q₂ = 9μC are placed 20 cm
apart. The electric field due to them will be zero on the line
joining them at a distance of
(a) 8 cm from q₁ (b) 8 cm form q₂
(c) 80/13 cm from q₁ (d) 80/13 cm from q₂

Answer 1

Given:

Two point charges q₁ = 4μC and q₂ = 9μC are placed 20 cm apart.

To find:

Position at which the net electrostatic field intensity is zero.

Calculation:

At the neutral point, electrostatic field intensity due to the two charges would cancel out each other.

Let that point be located x cm from 4 micro-C charge:

$\therefore \: E1 = E2$

$= > \dfrac{k(q1)}{ {x}^{2} } = \dfrac{k(q2)}{ {(20 - x)}^{2} }$

$= > \dfrac{k \times 4 \times {10}^{ - 6} }{ {x}^{2} } = \dfrac{k \times 9 \times {10}^{ - 6} }{ {(20 - x)}^{2} }$

$= > \dfrac{4 }{ {x}^{2} } = \dfrac{ 9 }{ {(20 - x)}^{2} }$

Taking square root on both sides:

$= > \dfrac{2}{x} = \dfrac{3}{20 - x}$

$= > 40 - 2x = 3x$

$= > 5x = 40$

$= > x = 8 \: cm$

So, that point is located 8 cm from q1.