Question

two point charges of 2C and 5C are 10 meters apart a third point charge of 6C is placed between then at what distance from 5C charge must the 6C charged is placed in such a way that it will stay in equilibrium

Answer 1

Answer:Answer : 2mfrom5C

Explanation : For E to be zero at a distance x from 5C

k5x2=k20(6−x)2

(6−x)2=4x2

6−x=2x

x=2m.

Answer 2

Answer:

The distance  between 5C and 6C charge should be 6.125m for it to remain in equilibrium.

Explanation:

Step 1:

6C charge is placed between the 2C and 5C such that  the distance between the 5C charge and 6C charge be x and distance between 2C and 6C be 10-x.

For equilibrium net force on 6C must be 0.

Step 2:

Force due to 5C charge,

F₁=$\frac{KQ1Q2}{x^{2} }$

F₁=$\frac{K*5*6}{x^{2} }$

Force due to 2C charge,

F₂=$\frac{KQ1Q2}{(10-x)^{2} }$

F₂=$\frac{K*2*6}{(10-x)^{2} }$

Step 3:

For Equilibrium F₁=F₂

$\frac{K*5*6}{x^{2} }$=$\frac{K*2*6}{(10-x)^{2} }$

On solving we get quadratic equation,

$3x^{2} -100x+500=0$

whose solutions are- 27.20 and 6.125.

Neglecting 27.20 the answer is 6.125.

The distance  between 5C and 6C charge should be 6.125m for it to remain in equilibrium.