Question

Charges q1 and q2 lie on the X axis at points X= -4 cm
and X=+4 cm respectively. How must q1 and q2 be related so that the net electrostatic force on a charge placed at X=+2 cm is zero?

I need a proper explanation, please​

Answer 1

Answer:

using Coulomb's law

F=Kq1q2/r^2

Kq1/6^2 =Kq2/2^2

q1=9q2

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Answer 2

Answer:

The charge q₁ is 9 times the charge q₂.

Explanation:

To solve this question we will use the following equation,

$F=\frac{kQq}{r^2}$      (1)

Where,

F=force acting between the charges

k=coulomb's constant

Q,q=charges between which the force is acting

r=distance between the charges

The force acting between q₁ and Q equation (1) can be written as,

$F_1=\frac{kQq_1}{6^2}$       (2)

r=2-(-4)=6cm

The force acting between q₂ and Q equation (1) can be written as,

$F_2=\frac{kQq_2}{2^2}$      (3)

r=4-2=2cm

Since the net force on charge Q is zero. Then,

$F_1=F_2$

$\frac{kQq_1}{6^2}=\frac{kQq_2}{2^2}$

$\frac{q_1}{36}=\frac{q_2}{4}$

$q_1=9q_2$

Hence, the charge q₁ is 9 times the charge q₂.