Question

two point charges 1 microcoulomb and 9 microcoulomb are placed at X is equals to minus 1 and X is equals to 3 on x-axis the position between two charges where net electric force on a unit list charge will be zero is

Answer 1

Answer:

x = 0

Explanation:

At such a point,electric field by 1μC charge and 9μC charge must be equal.

Using, F(electric) = k q/r²

  Here,

q₁ = 1uC  ;  q₂ = 9μC ;  r = |(-1)-(3)| = 4unit

  Let it that point be at a distance of x from -1. So at 4 - x from 3.

⇒ k (1μ)/x² = k(9μ)/(4 - x)²

⇒ 1/x² = 9/(4 - x)²

⇒ 1/x = ±3/(4 - x)

⇒ 4 - x = 3x   or  4 - x = -3

⇒ 1 = x         or  7 = x

 Hence, at distance of 1 from -1, means at x = 0, force on test charge would be 0

Answer 2

Answer:

$\large{\purple{\ddot{\smile}}}$

Answered by Rohith kumar maths dude

●Answer is x=0.

Explaination: -

●Using the formula F (electric)=kq/r^2

▪Here we know that q1=quC:q2=1not uCr=4unit .

▪Lets take a distance of x from -1so,

4-x from 3.

=k (1u not)/x^2=k (9unot)/(4-x)^2

=1/x^2=9/4-x^2

1/x=+/-3/(4-x).

▪We can get the values as

x=1 and x=7

▪Therefore, the distance of 1 ftom -1.

▪It means that force on that object would be zero.

☆Hope it helps u mate.

☆Thank you.