Question

2. Two point charges +1 nC and -4 nC are 1m apart in air. Find the positions along the line jpining the two charges at which resultant potential is zero.​

Answer 1

Answer:

1.33 m from -4 nC

Explanation:

Let x be the distance from -4nC, (1-x) be the distance from +1 nC.

k(+1nC)/(1-x) = k(-4nC)/x

-4+4x=x

4x-x=4

3x=4

x=4÷3=1.33 m

Answer 2

Answer:

The correct answer is 1.33 m from -4nC .

Explanation:

Given Data :

Two point charges +1 nC (q) and -4 nC (Q) which are 1 m apart.

Let at a distance x from -4 nC the resultant potential is zero.

That means, the distance from the +1 nC at which the potential will be zero is (1-x) .

Formula to be used :

$V = \frac{kq{}}{ {r} }$

Calculations:

At the point where the potential is zero,

The potentials of both the charges will be equal.

$\frac{k(1nC)}{(1 - x)} = \frac{k( - 4nC)}{x}$

$x = - 4(1 - x) \\ x = - 4 + 4x \\ x - 4x = - 4 \\ - 3x = - 4 \\ x = \frac{4}{3} = 1.33 \: m$

Hence, 1.33 m from the -4nC charge , the potential will be zero.

#SPJ2