Question

2 point charge 2nc and 18nc are located 1m apart air find the position along the line joining the 2charges at which the resultant electric intensity is zero​

Answer 1

Let the Electric intensity be zero at point P, at distance x from 2nC as shown in the figure attached.

So, at point P the electric intensity due to both charges must be equal and opposite.

Balancing the magnitude of electric intent due to them gives:

$\rm \implies \dfrac{ \cancel{k} \times 2}{ {x}^{2} } = \dfrac{ \cancel{k} \times 18}{ {(1 - x)}^{2} } \\ \\ \rm \implies {( \dfrac{1 - x}{x}) }^{2} = \dfrac{18}{2} \\ \\ \rm \implies {( \dfrac{1 - x}{x}) }^{2} = 9 \\ \\ \rm \implies ( \dfrac{1 - x}{x}) = \sqrt{9} \\ \\ \rm \implies ( \dfrac{1 - x}{x}) = 3 \\ \\ \rm \implies 1 - x = 3x \\ \\ \rm \implies 4x = 1 \\ \\ \rm \implies x = \frac{1}{4} \: m \\ \\ \rm \implies x = 0.25 \: m \\ \\ \rm \implies x = 25 \: cm$

$\therefore$ Resultant electric intensity will be zero at a distance 25 cm from 2nC and at 75 cm from 18nC

Answer 2

Question:-

2 point charge 2nc and 18nc are located 1m apart air find the position along the line joining the 2charges at which the resultant electric intensity is zero

Answer:-

Let the potential at point P is zero which is situated at a distance x to the left of 1nC charge as shown in the figure.

Potential due a charge q is given by V=kq/r

where k = 1/4€

Thus the potential at the point p

Vp = k (1nC) - k (4nC)

x 1 + x

But Vp = 0

➭ K (1nC) − k(4nC)

x 1 + x

Or

➭ 1/x - 4/1 - x = 0

Or

➭ 1/x = 4

1 + x

Or

➭ 3x = 1

➭ x = 0.33

☞ Hence verified